Class 12: Indefinite Integrals

$\displaystyle \text{Evaluate the following integrals:}$

$\displaystyle \textbf{Question 1: }~\int (x+1)\sqrt{x^2-x+1}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int (x+1)\sqrt{x^{2}-x+1}\,dx.$
$\displaystyle \text{Also, } x+1=\lambda\frac{d}{dx}(x^{2}-x+1)+\mu.$
$\displaystyle \Rightarrow x+1=\lambda(2x-1)+\mu.$
$\displaystyle \Rightarrow x+1=2\lambda x+\mu-\lambda.$
$\displaystyle \text{Equating coefficients of like terms,}$
$\displaystyle 2\lambda=1.$
$\displaystyle \Rightarrow \lambda=\frac{1}{2}.$
$\displaystyle \text{And}$
$\displaystyle \mu-\lambda=1.$
$\displaystyle \Rightarrow \mu-\frac{1}{2}=1.$
$\displaystyle \Rightarrow \mu=\frac{3}{2}.$
$\displaystyle \therefore I=\int\left[\frac{1}{2}(2x-1)+\frac{3}{2}\right]\sqrt{x^{2}-x+1}\,dx.$
$\displaystyle =\frac{1}{2}\int(2x-1)\sqrt{x^{2}-x+1}\,dx+\frac{3}{2}\int\sqrt{x^{2}-x+1}\,dx.$
$\displaystyle =\frac{1}{2}\int(2x-1)\sqrt{x^{2}-x+1}\,dx+\frac{3}{2}\int\sqrt{x^{2}-x+\frac{1}{4}-\frac{1}{4}+1}\,dx.$
$\displaystyle =\frac{1}{2}\int(2x-1)\sqrt{x^{2}-x+1}\,dx+\frac{3}{2}\int\sqrt{\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4}}\,dx.$
$\displaystyle =\frac{1}{2}\int(2x-1)\sqrt{x^{2}-x+1}\,dx+\frac{3}{2}\int\sqrt{\left(x-\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\,dx.$
$\displaystyle \text{Let } x^{2}-x+1=t.$
$\displaystyle \Rightarrow (2x-1)\,dx=dt.$
$\displaystyle \therefore I=\frac{1}{2}\int \sqrt{t}\,dt+\frac{3}{2}\int\sqrt{\left(x-\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\,dx.$
$\displaystyle =\frac{1}{2}\left(\frac{2}{3}t^{3/2}\right)+\frac{3}{2}\left[\frac{x-\frac{1}{2}}{2}\sqrt{\left(x-\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}+\frac{3}{8}\log\!\left|x-\frac{1}{2}+\sqrt{\left(x-\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\right|\right]+C.$
$\displaystyle =\frac{1}{3}(x^{2}-x+1)^{3/2}+\frac{3}{8}(2x-1)\sqrt{x^{2}-x+1}+\frac{9}{16}\log\!\left|x-\frac{1}{2}+\sqrt{x^{2}-x+1}\right|+C.$

$\displaystyle \textbf{Question 2: }~\int (x+1)\sqrt{2x^2+3}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int (x+1)\sqrt{2x^{2}+3}\,dx.$
$\displaystyle \text{Also, } x+1=\lambda\frac{d}{dx}(2x^{2}+3)+\mu.$
$\displaystyle \Rightarrow x+1=\lambda(4x)+\mu.$
$\displaystyle \text{Equating coefficients of like terms,}$
$\displaystyle 4\lambda=1.$
$\displaystyle \Rightarrow \lambda=\frac{1}{4} \text{ and } \mu=1.$
$\displaystyle \therefore I=\int\left[\frac{1}{4}(4x)+1\right]\sqrt{2x^{2}+3}\,dx.$
$\displaystyle =\frac{1}{4}\int(4x)\sqrt{2x^{2}+3}\,dx+\int\sqrt{2x^{2}+3}\,dx.$
$\displaystyle =\frac{1}{4}\int(4x)\sqrt{2x^{2}+3}\,dx+\int\sqrt{2\left(x^{2}+\frac{3}{2}\right)}\,dx.$
$\displaystyle =\frac{1}{4}\int(4x)\sqrt{2x^{2}+3}\,dx+\sqrt{2}\int\sqrt{x^{2}+\left(\frac{\sqrt{3}}{\sqrt{2}}\right)^{2}}\,dx.$
$\displaystyle \text{Let } 2x^{2}+3=t.$
$\displaystyle \Rightarrow 4x\,dx=dt.$
$\displaystyle \therefore I=\frac{1}{4}\int \sqrt{t}\,dt+\sqrt{2}\int\sqrt{x^{2}+\left(\frac{\sqrt{3}}{\sqrt{2}}\right)^{2}}\,dx.$
$\displaystyle =\frac{1}{4}\left(\frac{2}{3}t^{3/2}\right)+\sqrt{2}\left[\frac{x}{2}\sqrt{x^{2}+\frac{3}{2}}+\frac{3}{4}\ln\!\left|x+\sqrt{x^{2}+\frac{3}{2}}\right|\right]+C.$
$\displaystyle =\frac{1}{6}(2x^{2}+3)^{3/2}+\frac{x}{2}\sqrt{2x^{2}+3}+\frac{3\sqrt{2}}{4}\log\!\left|x+\frac{\sqrt{2x^{2}+3}}{\sqrt{2}}\right|+C.$
$\displaystyle =\frac{1}{6}(2x^{2}+3)^{3/2}+\frac{x}{2}\sqrt{2x^{2}+3}+\frac{3\sqrt{2}}{4}\log\!\left|\sqrt{2}x+\sqrt{2x^{2}+3}\right|+C'.$
$\displaystyle \text{where } C'=C-\frac{3\sqrt{2}}{4}\log\sqrt{2}.$

$\displaystyle \textbf{Question 3: }~\int (2x-5)\sqrt{2+3x-x^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int (2x-5)\sqrt{2+3x-x^{2}}\,dx.$
$\displaystyle \text{Also, } 2x-5=\lambda\frac{d}{dx}(2+3x-x^{2})+\mu.$
$\displaystyle \Rightarrow 2x-5=\lambda(-2x+3)+\mu.$
$\displaystyle \Rightarrow 2x-5=-2\lambda x+3\lambda+\mu.$
$\displaystyle \text{Equating coefficients of like terms,}$
$\displaystyle -2\lambda=2.$
$\displaystyle \Rightarrow \lambda=-1.$
$\displaystyle \text{And}$
$\displaystyle 3\lambda+\mu=-5.$
$\displaystyle \Rightarrow 3(-1)+\mu=-5.$
$\displaystyle \Rightarrow \mu=-2.$
$\displaystyle \therefore 2x-5=-1(-2x+3)-2.$
$\displaystyle \text{Hence, } I=\int\left[-(-2x+3)-2\right]\sqrt{2+3x-x^{2}}\,dx.$
$\displaystyle =-\int(-2x+3)\sqrt{2+3x-x^{2}}\,dx-2\int\sqrt{2+3x-x^{2}}\,dx.$
$\displaystyle =-I_1-2I_2 \quad \ldots (1).$
$\displaystyle \text{where } I_1=\int(-2x+3)\sqrt{2+3x-x^{2}}\,dx.$
$\displaystyle \text{Let } 2+3x-x^{2}=t.$
$\displaystyle \Rightarrow (-2x+3)\,dx=dt.$
$\displaystyle \therefore I_1=\int t^{1/2}\,dt.$
$\displaystyle I_1=\frac{2}{3}t^{3/2}.$
$\displaystyle I_1=\frac{2}{3}(2+3x-x^{2})^{3/2} \quad \ldots (2).$
$\displaystyle \text{And } I_2=\int\sqrt{2+3x-x^{2}}\,dx.$
$\displaystyle I_2=\int\sqrt{2-(x^{2}-3x)}\,dx.$
$\displaystyle =\int\sqrt{2-\left(x^{2}-3x+\frac{9}{4}-\frac{9}{4}\right)}\,dx.$
$\displaystyle =\int\sqrt{\frac{17}{4}-\left(x-\frac{3}{2}\right)^{2}}\,dx.$
$\displaystyle =\int\sqrt{\left(\frac{\sqrt{17}}{2}\right)^{2}-\left(x-\frac{3}{2}\right)^{2}}\,dx.$
$\displaystyle =\frac{x-\frac{3}{2}}{2}\sqrt{\left(\frac{\sqrt{17}}{2}\right)^{2}-\left(x-\frac{3}{2}\right)^{2}}+\frac{\left(\frac{\sqrt{17}}{2}\right)^{2}}{2}\sin^{-1}\!\left(\frac{x-\frac{3}{2}}{\frac{\sqrt{17}}{2}}\right)+C.$
$\displaystyle =\frac{2x-3}{4}\sqrt{2+3x-x^{2}}+\frac{17}{8}\sin^{-1}\!\left(\frac{2x-3}{\sqrt{17}}\right) \quad \ldots (3).$
$\displaystyle \text{From (1), (2) and (3),}$
$\displaystyle I=-\frac{2}{3}(2+3x-x^{2})^{3/2}-\frac{2x-3}{2}\sqrt{2+3x-x^{2}}-\frac{17}{4}\sin^{-1}\!\left(\frac{2x-3}{\sqrt{17}}\right)+C.$

$\displaystyle \textbf{Question 4: }~\int (x+2)\sqrt{x^2+x+1}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int (x+2)\sqrt{x^{2}+x+1}\,dx.$
$\displaystyle \text{Also, } x+2=\lambda\frac{d}{dx}(x^{2}+x+1)+\mu.$
$\displaystyle \Rightarrow x+2=\lambda(2x+1)+\mu.$
$\displaystyle \Rightarrow x+2=2\lambda x+\lambda+\mu.$
$\displaystyle \text{Equating coefficients of like terms,}$
$\displaystyle 2\lambda=1.$
$\displaystyle \Rightarrow \lambda=\frac{1}{2}.$
$\displaystyle \text{And } \lambda+\mu=2.$
$\displaystyle \Rightarrow \frac{1}{2}+\mu=2.$
$\displaystyle \Rightarrow \mu=\frac{3}{2}.$
$\displaystyle \therefore I=\int\left[\frac{1}{2}(2x+1)+\frac{3}{2}\right]\sqrt{x^{2}+x+1}\,dx.$
$\displaystyle =\frac{1}{2}\int(2x+1)\sqrt{x^{2}+x+1}\,dx+\frac{3}{2}\int\sqrt{x^{2}+x+1}\,dx.$
$\displaystyle =\frac{1}{2}\int(2x+1)\sqrt{x^{2}+x+1}\,dx+\frac{3}{2}\int\sqrt{x^{2}+x+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+1}\,dx.$
$\displaystyle =\frac{1}{2}\int(2x+1)\sqrt{x^{2}+x+1}\,dx+\frac{3}{2}\int\sqrt{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\,dx.$
$\displaystyle \text{Let } x^{2}+x+1=t.$
$\displaystyle \Rightarrow (2x+1)\,dx=dt.$
$\displaystyle \therefore I=\frac{1}{2}\int t^{1/2}\,dt+\frac{3}{2}\int\sqrt{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\,dx.$
$\displaystyle =\frac{1}{2}\left(\frac{2}{3}t^{3/2}\right)+\frac{3}{2}\left[\frac{x+\frac{1}{2}}{2}\sqrt{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}+\frac{3}{8}\log\!\left|x+\frac{1}{2}+\sqrt{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\right|\right]+C.$
$\displaystyle =\frac{1}{3}(x^{2}+x+1)^{3/2}+\frac{3}{8}(2x+1)\sqrt{x^{2}+x+1}+\frac{9}{16}\log\!\left|x+\frac{1}{2}+\sqrt{x^{2}+x+1}\right|+C.$

$\displaystyle \textbf{Question 5: }~\int (4x+1)\sqrt{x^2-x-2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int (4x+1)\sqrt{x^{2}-x-2}\,dx.$
$\displaystyle \text{Also, } 4x+1=\lambda\frac{d}{dx}(x^{2}-x-2)+\mu.$
$\displaystyle \Rightarrow 4x+1=\lambda(2x-1)+\mu.$
$\displaystyle \Rightarrow 4x+1=2\lambda x+\mu-\lambda.$
$\displaystyle \text{Equating coefficients of like terms,}$
$\displaystyle 2\lambda=4.$
$\displaystyle \Rightarrow \lambda=2.$
$\displaystyle \text{And } \mu-\lambda=1.$
$\displaystyle \Rightarrow \mu-2=1.$
$\displaystyle \Rightarrow \mu=3.$
$\displaystyle \therefore I=\int\left[2(2x-1)+3\right]\sqrt{x^{2}-x-2}\,dx.$
$\displaystyle =2\int(2x-1)\sqrt{x^{2}-x-2}\,dx+3\int\sqrt{x^{2}-x-2}\,dx.$
$\displaystyle =2\int(2x-1)\sqrt{x^{2}-x-2}\,dx+3\int\sqrt{x^{2}-x+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}-2}\,dx.$
$\displaystyle =2\int(2x-1)\sqrt{x^{2}-x-2}\,dx+3\int\sqrt{\left(x-\frac{1}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}}\,dx.$
$\displaystyle \text{Let } x^{2}-x-2=t.$
$\displaystyle \Rightarrow (2x-1)\,dx=dt.$
$\displaystyle \therefore I=2\int\sqrt{t}\,dt+3\int\sqrt{\left(x-\frac{1}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}}\,dx.$
$\displaystyle =2\left(\frac{2}{3}t^{3/2}\right)+3\left[\frac{x-\frac{1}{2}}{2}\sqrt{\left(x-\frac{1}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}}-\frac{\left(\frac{3}{2}\right)^{2}}{2}\log\!\left|x-\frac{1}{2}+\sqrt{x^{2}-x-2}\right|\right]+C.$
$\displaystyle =\frac{4}{3}(x^{2}-x-2)^{3/2}+\frac{3}{4}(2x-1)\sqrt{x^{2}-x-2}-\frac{27}{8}\log\!\left|x-\frac{1}{2}+\sqrt{x^{2}-x-2}\right|+C.$

$\displaystyle \textbf{Question 6: }~\int (x-2)\sqrt{2x^2-6x+5}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int (x-2)\sqrt{2x^{2}-6x+5}\,dx.$
$\displaystyle \text{Also, } x-2=\lambda\frac{d}{dx}(2x^{2}-6x+5)+\mu.$
$\displaystyle \Rightarrow x-2=\lambda(4x-6)+\mu.$
$\displaystyle \Rightarrow x-2=4\lambda x+\mu-6\lambda.$
$\displaystyle \text{Equating coefficients of like terms,}$
$\displaystyle 4\lambda=1.$
$\displaystyle \Rightarrow \lambda=\frac{1}{4}.$
$\displaystyle \text{And } \mu-6\lambda=-2.$
$\displaystyle \Rightarrow \mu-\frac{6}{4}=-2.$
$\displaystyle \Rightarrow \mu=-\frac{1}{2}.$
$\displaystyle \therefore I=\int\left[\frac{1}{4}(4x-6)-\frac{1}{2}\right]\sqrt{2x^{2}-6x+5}\,dx.$
$\displaystyle =\frac{1}{4}\int(4x-6)\sqrt{2x^{2}-6x+5}\,dx-\frac{1}{2}\int\sqrt{2x^{2}-6x+5}\,dx.$
$\displaystyle \text{Let } 2x^{2}-6x+5=t.$
$\displaystyle \Rightarrow (4x-6)\,dx=dt.$
$\displaystyle \therefore I=\frac{1}{4}\int t^{1/2}\,dt-\frac{1}{2}\int\sqrt{2\left(x^{2}-3x+\frac{5}{2}\right)}\,dx.$
$\displaystyle =\frac{1}{4}\int t^{1/2}\,dt-\frac{\sqrt{2}}{2}\int\sqrt{x^{2}-3x+\frac{9}{4}-\frac{9}{4}+\frac{5}{2}}\,dx.$
$\displaystyle =\frac{1}{4}\int t^{1/2}\,dt-\frac{\sqrt{2}}{2}\int\sqrt{\left(x-\frac{3}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}}\,dx.$
$\displaystyle =\frac{1}{6}t^{3/2}-\frac{1}{\sqrt{2}}\left[\frac{x-\frac{3}{2}}{2}\sqrt{\left(x-\frac{3}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}}+\frac{1}{8}\log\!\left|x-\frac{3}{2}+\sqrt{x^{2}-3x+\frac{5}{2}}\right|\right]+C.$
$\displaystyle =\frac{1}{6}(2x^{2}-6x+5)^{3/2}-\frac{2x-3}{4\sqrt{2}}\sqrt{x^{2}-3x+\frac{5}{2}}-\frac{1}{8\sqrt{2}}\log\!\left|\frac{2x-3}{2}+\sqrt{x^{2}-3x+\frac{5}{2}}\right|+C.$

$\displaystyle \textbf{Question 7: }~\int (x+1)\sqrt{x^2+x+1}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int (x+1)\sqrt{x^{2}+x+1}\,dx.$
$\displaystyle \text{Also, } x+1=\lambda\frac{d}{dx}(x^{2}+x+1)+\mu.$
$\displaystyle \Rightarrow x+1=\lambda(2x+1)+\mu.$
$\displaystyle \Rightarrow x+1=2\lambda x+\lambda+\mu.$
$\displaystyle \text{Equating coefficients of like terms,}$
$\displaystyle 2\lambda=1.$
$\displaystyle \Rightarrow \lambda=\frac{1}{2}.$
$\displaystyle \text{And } \lambda+\mu=1.$
$\displaystyle \Rightarrow \frac{1}{2}+\mu=1.$
$\displaystyle \Rightarrow \mu=\frac{1}{2}.$
$\displaystyle \therefore I=\frac{1}{2}\int(2x+1)\sqrt{x^{2}+x+1}\,dx+\frac{1}{2}\int\sqrt{x^{2}+x+1}\,dx.$
$\displaystyle =\frac{1}{2}\int(2x+1)\sqrt{x^{2}+x+1}\,dx+\frac{1}{2}\int\sqrt{x^{2}+x+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+1}\,dx.$
$\displaystyle =\frac{1}{2}\int(2x+1)\sqrt{x^{2}+x+1}\,dx+\frac{1}{2}\int\sqrt{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\,dx.$
$\displaystyle \text{Let } x^{2}+x+1=t.$
$\displaystyle \Rightarrow (2x+1)\,dx=dt.$
$\displaystyle \therefore I=\frac{1}{2}\int t^{1/2}\,dt+\frac{1}{2}\int\sqrt{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\,dx.$
$\displaystyle =\frac{1}{2}\left(\frac{2}{3}t^{3/2}\right)+\frac{1}{2}\left[\frac{x+\frac{1}{2}}{2}\sqrt{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}+\frac{3}{8}\log\!\left|x+\frac{1}{2}+\sqrt{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\right|\right]+C.$
$\displaystyle =\frac{1}{3}(x^{2}+x+1)^{3/2}+\frac{1}{2}\left[\frac{2x+1}{4}\sqrt{x^{2}+x+1}+\frac{3}{8}\log\!\left|x+\frac{1}{2}+\sqrt{x^{2}+x+1}\right|\right]+C.$

$\displaystyle \textbf{Question 8: }~\int (2x+3)\sqrt{x^2+4x+3}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int (2x+3)\sqrt{x^{2}+4x+3}\,dx.$
$\displaystyle \text{Also, } 2x+3=\lambda\frac{d}{dx}(x^{2}+4x+3)+\mu.$
$\displaystyle \Rightarrow 2x+3=\lambda(2x+4)+\mu.$
$\displaystyle \Rightarrow 2x+3=2\lambda x+4\lambda+\mu.$
$\displaystyle \text{Equating coefficients of like terms,}$
$\displaystyle 2\lambda=2.$
$\displaystyle \Rightarrow \lambda=1.$
$\displaystyle \text{And } 4\lambda+\mu=3.$
$\displaystyle \Rightarrow 4+\mu=3.$
$\displaystyle \Rightarrow \mu=-1.$
$\displaystyle \therefore I=\int\left[(2x+4)-1\right]\sqrt{x^{2}+4x+3}\,dx.$
$\displaystyle =\int(2x+4)\sqrt{x^{2}+4x+3}\,dx-\int\sqrt{x^{2}+4x+3}\,dx.$
$\displaystyle =\int(2x+4)\sqrt{x^{2}+4x+3}\,dx-\int\sqrt{x^{2}+4x+4-1}\,dx.$
$\displaystyle =\int(2x+4)\sqrt{x^{2}+4x+3}\,dx-\int\sqrt{(x+2)^{2}-1}\,dx.$
$\displaystyle \text{Let } x^{2}+4x+3=t.$
$\displaystyle \Rightarrow (2x+4)\,dx=dt.$
$\displaystyle \therefore I=\int\sqrt{t}\,dt-\int\sqrt{(x+2)^{2}-1}\,dx.$
$\displaystyle =\frac{2}{3}t^{3/2}-\left[\frac{x+2}{2}\sqrt{(x+2)^{2}-1}-\frac{1}{2}\log\!\left|(x+2)+\sqrt{(x+2)^{2}-1}\right|\right]+C.$
$\displaystyle =\frac{2}{3}(x^{2}+4x+3)^{3/2}-\frac{x+2}{2}\sqrt{x^{2}+4x+3}+\frac{1}{2}\log\!\left|(x+2)+\sqrt{x^{2}+4x+3}\right|+C.$

$\displaystyle \textbf{Question 9: }~\int (2x-5)\sqrt{x^2-4x+3}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int (2x-5)\sqrt{x^{2}-4x+3}\,dx.$
$\displaystyle =\int \big[(2x-4)-1\big]\sqrt{x^{2}-4x+3}\,dx.$
$\displaystyle =\int (2x-4)\sqrt{x^{2}-4x+3}\,dx-\int \sqrt{x^{2}-4x+3}\,dx.$
$\displaystyle =\int (2x-4)\sqrt{x^{2}-4x+3}\,dx-\int \sqrt{x^{2}-4x+4-1}\,dx.$
$\displaystyle =\int (2x-4)\sqrt{x^{2}-4x+3}\,dx-\int \sqrt{(x-2)^{2}-1^{2}}\,dx.$
$\displaystyle \text{Let } x^{2}-4x+3=t.$
$\displaystyle \Rightarrow (2x-4)\,dx=dt.$
$\displaystyle \therefore I=\int \sqrt{t}\,dt-\int \sqrt{(x-2)^{2}-1^{2}}\,dx.$
$\displaystyle =\frac{2}{3}t^{3/2}-\left[\frac{x-2}{2}\sqrt{(x-2)^{2}-1}-\frac{1}{2}\log\!\left|(x-2)+\sqrt{(x-2)^{2}-1}\right|\right]+C.$
$\displaystyle =\frac{2}{3}(x^{2}-4x+3)^{3/2}-\frac{x-2}{2}\sqrt{x^{2}-4x+3}+\frac{1}{2}\log\!\left|(x-2)+\sqrt{x^{2}-4x+3}\right|+C.$

$\displaystyle \textbf{Question 10: }~\int x\sqrt{x^2+x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int x\sqrt{x^{2}+x}\,dx.$
$\displaystyle \text{Also, } x=\lambda\frac{d}{dx}(x^{2}+x)+\mu.$
$\displaystyle \Rightarrow x=\lambda(2x+1)+\mu.$
$\displaystyle \Rightarrow x=2\lambda x+\lambda+\mu.$
$\displaystyle \text{Equating coefficients of like terms,}$
$\displaystyle 2\lambda=1.$
$\displaystyle \Rightarrow \lambda=\frac{1}{2}.$
$\displaystyle \text{And } \lambda+\mu=0.$
$\displaystyle \Rightarrow \mu=-\frac{1}{2}.$
$\displaystyle \therefore I=\int\left[\frac{1}{2}(2x+1)-\frac{1}{2}\right]\sqrt{x^{2}+x}\,dx.$
$\displaystyle =\frac{1}{2}\int(2x+1)\sqrt{x^{2}+x}\,dx-\frac{1}{2}\int\sqrt{x^{2}+x}\,dx.$
$\displaystyle =\frac{1}{2}\int(2x+1)\sqrt{x^{2}+x}\,dx-\frac{1}{2}\int\sqrt{x^{2}+x+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}\,dx.$
$\displaystyle =\frac{1}{2}\int(2x+1)\sqrt{x^{2}+x}\,dx-\frac{1}{2}\int\sqrt{\left(x+\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}\,dx.$
$\displaystyle \text{Let } x^{2}+x=t.$
$\displaystyle \Rightarrow (2x+1)\,dx=dt.$
$\displaystyle \therefore I=\frac{1}{2}\int t^{1/2}\,dt-\frac{1}{2}\int\sqrt{\left(x+\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}\,dx.$
$\displaystyle =\frac{1}{2}\left(\frac{2}{3}t^{3/2}\right)-\frac{1}{2}\left[\frac{x+\frac{1}{2}}{2}\sqrt{\left(x+\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}-\frac{1}{8}\log\!\left|x+\frac{1}{2}+\sqrt{x^{2}+x}\right|\right]+C.$
$\displaystyle =\frac{1}{3}(x^{2}+x)^{3/2}-\frac{2x+1}{8}\sqrt{x^{2}+x}+\frac{1}{16}\log\!\left|x+\frac{1}{2}+\sqrt{x^{2}+x}\right|+C.$

$\displaystyle \textbf{Question 11: }~\int (x-3)\sqrt{x^2+3x-18}\,dx. \hspace{6.0cm} \text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int (x-3)\sqrt{x^{2}+3x-18}\,dx.$
$\displaystyle \text{We express } x-3=A\frac{d}{dx}(x^{2}+3x-18)+B.$
$\displaystyle \Rightarrow x-3=A(2x+3)+B.$
$\displaystyle \text{Equating coefficients of } x \text{ and constants,}$
$\displaystyle 1=2A \text{ and } -3=3A+B.$
$\displaystyle \Rightarrow A=\frac{1}{2} \text{ and } B=-\frac{9}{2}.$
$\displaystyle \therefore I=\int\left[\frac{1}{2}(2x+3)-\frac{9}{2}\right]\sqrt{x^{2}+3x-18}\,dx.$
$\displaystyle =\frac{1}{2}\int(2x+3)\sqrt{x^{2}+3x-18}\,dx-\frac{9}{2}\int\sqrt{x^{2}+3x-18}\,dx.$
$\displaystyle =\frac{1}{2}I_1-\frac{9}{2}I_2 \quad \ldots (1).$
$\displaystyle \text{Now, } I_1=\int(2x+3)\sqrt{x^{2}+3x-18}\,dx.$
$\displaystyle \text{Let } x^{2}+3x-18=u.$
$\displaystyle \Rightarrow (2x+3)\,dx=du.$
$\displaystyle \therefore I_1=\int \sqrt{u}\,du.$
$\displaystyle I_1=\frac{2}{3}u^{3/2}+c_1.$
$\displaystyle I_1=\frac{2}{3}(x^{2}+3x-18)^{3/2}+c_1 \quad \ldots (2).$
$\displaystyle \text{And } I_2=\int\sqrt{x^{2}+3x-18}\,dx.$
$\displaystyle =\int\sqrt{x^{2}+3x+\frac{9}{4}-\frac{9}{4}-18}\,dx.$
$\displaystyle =\int\sqrt{\left(x+\frac{3}{2}\right)^{2}-\left(\frac{9}{2}\right)^{2}}\,dx.$
$\displaystyle \text{Let } \left(x+\frac{3}{2}\right)=w.$
$\displaystyle \Rightarrow dx=dw.$
$\displaystyle \therefore I_2=\int\sqrt{w^{2}-\left(\frac{9}{2}\right)^{2}}\,dw.$
$\displaystyle I_2=\frac{w}{2}\sqrt{w^{2}-\left(\frac{9}{2}\right)^{2}}-\frac{\left(\frac{9}{2}\right)^{2}}{2}\log\!\left|w+\sqrt{w^{2}-\left(\frac{9}{2}\right)^{2}}\right|+c_2.$
$\displaystyle I_2=\frac{2x+3}{4}\sqrt{x^{2}+3x-18}-\frac{81}{8}\log\!\left|x+\frac{3}{2}+\sqrt{x^{2}+3x-18}\right|+c_2 \quad \ldots (3).$
$\displaystyle \text{From (1), (2) and (3),}$
$\displaystyle I=\frac{1}{3}(x^{2}+3x-18)^{3/2}-\frac{9}{8}(2x+3)\sqrt{x^{2}+3x-18}+\frac{729}{16}\log\!\left|x+\frac{3}{2}+\sqrt{x^{2}+3x-18}\right|+C.$
$\displaystyle \text{Hence, } \int (x-3)\sqrt{x^{2}+3x-18}\,dx=\frac{1}{3}(x^{2}+3x-18)^{3/2}-\frac{9}{8}(2x+3)\sqrt{x^{2}+3x-18}+\frac{729}{16}\log\!\left|x+\frac{3}{2}+\sqrt{x^{2}+3x-18}\right|+C.$

$\displaystyle \textbf{Question 12: }~\int (x+3)\sqrt{3-4x-x^2}\,dx. \hspace{6.0cm} \text{[CBSE 2014, 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int (x+3)\sqrt{3-4x-x^{2}}\,dx.$
$\displaystyle \text{We express } x+3=A\frac{d}{dx}(3-4x-x^{2})+B.$
$\displaystyle \Rightarrow x+3=A(-4-2x)+B.$
$\displaystyle \text{Equating coefficients of } x \text{ and constants, we get}$
$\displaystyle 1=-2A \text{ and } 3=-4A+B.$
$\displaystyle \Rightarrow A=-\frac{1}{2} \text{ and } B=1.$
$\displaystyle \therefore I=\int\left[-\frac{1}{2}(-4-2x)+1\right]\sqrt{3-4x-x^{2}}\,dx.$
$\displaystyle =-\frac{1}{2}\int(-4-2x)\sqrt{3-4x-x^{2}}\,dx+\int\sqrt{3-4x-x^{2}}\,dx.$
$\displaystyle =-\frac{1}{2}I_1+I_2 \quad \ldots (1).$
$\displaystyle \text{Now, } I_1=\int(-4-2x)\sqrt{3-4x-x^{2}}\,dx.$
$\displaystyle \text{Let } 3-4x-x^{2}=u.$
$\displaystyle \Rightarrow (-4-2x)\,dx=du.$
$\displaystyle \therefore I_1=\int \sqrt{u}\,du.$
$\displaystyle I_1=\frac{2}{3}u^{3/2}+c_1.$
$\displaystyle I_1=\frac{2}{3}(3-4x-x^{2})^{3/2}+c_1 \quad \ldots (2).$
$\displaystyle \text{And, } I_2=\int\sqrt{3-4x-x^{2}}\,dx.$
$\displaystyle =\int\sqrt{7-(x+2)^{2}}\,dx.$
$\displaystyle \text{Let } (x+2)=u.$
$\displaystyle \Rightarrow dx=du.$
$\displaystyle \therefore I_2=\int\sqrt{(\sqrt{7})^{2}-u^{2}}\,du.$
$\displaystyle I_2=\frac{u}{2}\sqrt{(\sqrt{7})^{2}-u^{2}}+\frac{(\sqrt{7})^{2}}{2}\sin^{-1}\!\left(\frac{u}{\sqrt{7}}\right)+c_2.$
$\displaystyle I_2=\frac{x+2}{2}\sqrt{3-4x-x^{2}}+\frac{7}{2}\sin^{-1}\!\left(\frac{x+2}{\sqrt{7}}\right)+c_2 \quad \ldots (3).$
$\displaystyle \text{From (1), (2) and (3), we get}$
$\displaystyle I=-\frac{1}{3}(3-4x-x^{2})^{3/2}+\frac{x+2}{2}\sqrt{3-4x-x^{2}}+\frac{7}{2}\sin^{-1}\!\left(\frac{x+2}{\sqrt{7}}\right)+C.$
$\displaystyle \text{Hence, } \int (x+3)\sqrt{3-4x-x^{2}}\,dx=-\frac{1}{3}(3-4x-x^{2})^{3/2}+\frac{x+2}{2}\sqrt{3-4x-x^{2}}+\frac{7}{2}\sin^{-1}\!\left(\frac{x+2}{\sqrt{7}}\right)+C.$

$\displaystyle \textbf{Question 13: }~\int (3x+1)\sqrt{4-3x-2x^2}\,dx. \hspace{6.0cm} \text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int(3x+1)\sqrt{4-3x-2x^{2}}\,dx.$
$\displaystyle \text{Let } (3x+1)=A\frac{d}{dx}(4-3x-2x^{2})+B.$
$\displaystyle \Rightarrow (3x+1)=A(-3-4x)+B.$
$\displaystyle \Rightarrow (3x+1)=-4Ax+(B-3A).$
$\displaystyle \text{Equating coefficients of like terms,}$
$\displaystyle -4A=3 \text{ and } B-3A=1.$
$\displaystyle \Rightarrow A=-\frac{3}{4} \text{ and } B=-\frac{5}{4}.$
$\displaystyle \therefore I=\int\left[-\frac{3}{4}(-3-4x)-\frac{5}{4}\right]\sqrt{4-3x-2x^{2}}\,dx.$
$\displaystyle =-\frac{3}{4}\int(-3-4x)\sqrt{4-3x-2x^{2}}\,dx-\frac{5}{4}\int\sqrt{4-3x-2x^{2}}\,dx.$
$\displaystyle \text{Let } I=-\frac{3}{4}I_1-\frac{5}{4}I_2 \quad \ldots (i).$
$\displaystyle \text{Now, } I_1=\int(-3-4x)\sqrt{4-3x-2x^{2}}\,dx.$
$\displaystyle \text{Let } (4-3x-2x^{2})=t.$
$\displaystyle \Rightarrow (-3-4x)\,dx=dt.$
$\displaystyle \therefore I_1=\int \sqrt{t}\,dt.$
$\displaystyle I_1=\frac{2}{3}t^{3/2}+c_1.$
$\displaystyle I_1=\frac{2}{3}(4-3x-2x^{2})^{3/2}+c_1.$
$\displaystyle \text{And } I_2=\int\sqrt{4-3x-2x^{2}}\,dx.$
$\displaystyle =\int\sqrt{2\left(2-\frac{3}{2}x-x^{2}\right)}\,dx.$
$\displaystyle =\sqrt{2}\int\sqrt{\frac{17}{4}-\left(x+\frac{3}{4}\right)^{2}}\,dx.$
$\displaystyle \text{Let } \left(x+\frac{3}{4}\right)=u.$
$\displaystyle \Rightarrow dx=du.$
$\displaystyle \therefore I_2=\sqrt{2}\int\sqrt{\left(\frac{\sqrt{17}}{2}\right)^{2}-u^{2}}\,du.$
$\displaystyle I_2=\sqrt{2}\left[\frac{u}{2}\sqrt{\left(\frac{\sqrt{17}}{2}\right)^{2}-u^{2}}+\frac{\left(\frac{\sqrt{17}}{2}\right)^{2}}{2}\sin^{-1}\!\left(\frac{u}{\frac{\sqrt{17}}{2}}\right)\right]+c_2.$
$\displaystyle I_2=\frac{x+\frac{3}{4}}{\sqrt{2}}\sqrt{4-3x-2x^{2}}+\frac{17}{4\sqrt{2}}\sin^{-1}\!\left(\frac{2x+3}{\sqrt{17}}\right)+c_2.$
$\displaystyle \text{Using (i), we get}$
$\displaystyle I=-\frac{1}{2}(4-3x-2x^{2})^{3/2}-\frac{5}{4}\left[\frac{x+\frac{3}{4}}{\sqrt{2}}\sqrt{4-3x-2x^{2}}+\frac{17}{4\sqrt{2}}\sin^{-1}\!\left(\frac{2x+3}{\sqrt{17}}\right)\right]+C.$

$\displaystyle \textbf{Question 14: }~\int (2x+5)\sqrt{10-4x-3x^2}\,dx. \hspace{6.0cm} \text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int(2x+5)\sqrt{10-4x-3x^{2}}\,dx.$
$\displaystyle \text{Let } (2x+5)=A\frac{d}{dx}(10-4x-3x^{2})+B.$
$\displaystyle \Rightarrow (2x+5)=A(-4-6x)+B.$
$\displaystyle \Rightarrow (2x+5)=-6Ax+(B-4A).$
$\displaystyle \text{Equating coefficients of like terms,}$
$\displaystyle -6A=2 \text{ and } B-4A=5.$
$\displaystyle \Rightarrow A=-\frac{1}{3} \text{ and } B=\frac{11}{3}.$
$\displaystyle \therefore I=\int\left[-\frac{1}{3}(-4-6x)+\frac{11}{3}\right]\sqrt{10-4x-3x^{2}}\,dx.$
$\displaystyle =-\frac{1}{3}\int(-4-6x)\sqrt{10-4x-3x^{2}}\,dx+\frac{11}{3}\int\sqrt{10-4x-3x^{2}}\,dx.$
$\displaystyle \text{Let } I=-\frac{1}{3}I_1+\frac{11}{3}I_2 \quad \ldots (i).$
$\displaystyle \text{Now, } I_1=\int(-4-6x)\sqrt{10-4x-3x^{2}}\,dx.$
$\displaystyle \text{Let } (10-4x-3x^{2})=t.$
$\displaystyle \Rightarrow (-4-6x)\,dx=dt.$
$\displaystyle \therefore I_1=\int \sqrt{t}\,dt.$
$\displaystyle I_1=\frac{2}{3}t^{3/2}+c_1.$
$\displaystyle I_1=\frac{2}{3}(10-4x-3x^{2})^{3/2}+c_1.$
$\displaystyle \text{And } I_2=\int\sqrt{10-4x-3x^{2}}\,dx.$
$\displaystyle =\int\sqrt{3\left(\frac{10}{3}-\frac{4}{3}x-x^{2}\right)}\,dx.$
$\displaystyle =\sqrt{3}\int\sqrt{\frac{26}{9}-\left(x+\frac{2}{3}\right)^{2}}\,dx.$
$\displaystyle \text{Let } \left(x+\frac{2}{3}\right)=u.$
$\displaystyle \Rightarrow dx=du.$
$\displaystyle \therefore I_2=\sqrt{3}\int\sqrt{\left(\frac{\sqrt{26}}{3}\right)^{2}-u^{2}}\,du.$
$\displaystyle I_2=\sqrt{3}\left[\frac{u}{2}\sqrt{\left(\frac{\sqrt{26}}{3}\right)^{2}-u^{2}}+\frac{\left(\frac{\sqrt{26}}{3}\right)^{2}}{2}\sin^{-1}\!\left(\frac{u}{\frac{\sqrt{26}}{3}}\right)\right]+c_2.$
$\displaystyle I_2=\frac{\sqrt{3}}{2}\left(x+\frac{2}{3}\right)\sqrt{10-4x-3x^{2}}+\frac{13\sqrt{3}}{9}\sin^{-1}\!\left(\frac{3x+2}{\sqrt{26}}\right)+c_2.$
$\displaystyle \text{Using (i), we get}$
$\displaystyle I=-\frac{2}{9}(10-4x-3x^{2})^{3/2}+\frac{11\sqrt{3}}{6}\left(x+\frac{2}{3}\right)\sqrt{10-4x-3x^{2}}+\frac{143\sqrt{3}}{27}\sin^{-1}\!\left(\frac{3x+2}{\sqrt{26}}\right)+C.$