\displaystyle \textbf{Evaluate the following integrals:}

\displaystyle \textbf{Question 1: }~\int \frac{x}{\sqrt{x^2+6x+10}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{x\,dx}{\sqrt{x^2+6x+10}}
\displaystyle x=A\frac{d}{dx}(x^2+6x+10)+B
\displaystyle x=A(2x+6)+B
\displaystyle x=(2A)x+6A+B
\displaystyle \text{Equating coefficients of like terms}
\displaystyle 2A=1
\displaystyle A=\frac12
\displaystyle 6A+B=0
\displaystyle 6\cdot\frac12+B=0
\displaystyle B=-3
\displaystyle I=\int \frac{\frac12(2x+6)-3}{\sqrt{x^2+6x+10}}\,dx
\displaystyle =\frac12\int \frac{2x+6}{\sqrt{x^2+6x+10}}\,dx-3\int \frac{dx}{\sqrt{x^2+6x+10}}
\displaystyle =\frac12\int \frac{2x+6}{\sqrt{x^2+6x+10}}\,dx-3\int \frac{dx}{\sqrt{x^2+6x+3^2-3^2+10}}
\displaystyle =\frac12\int \frac{2x+6}{\sqrt{x^2+6x+10}}\,dx-3\int \frac{dx}{\sqrt{(x+3)^2+1^2}}
\displaystyle \text{Let } x^2+6x+10=t
\displaystyle \Rightarrow (2x+6)\,dx=dt
\displaystyle I=\frac12\int \frac{dt}{\sqrt{t}}-3\int \frac{dx}{\sqrt{(x+3)^2+1}}
\displaystyle =\frac12\cdot2\sqrt{t}-3\log\left|x+3+\sqrt{(x+3)^2+1}\right|+C
\displaystyle =\sqrt{t}-3\log\left|x+3+\sqrt{x^2+6x+10}\right|+C
\displaystyle =\sqrt{x^2+6x+10}-3\log\left|x+3+\sqrt{x^2+6x+10}\right|+C

\displaystyle \textbf{Question 2: }~\int \frac{2x+1}{\sqrt{x^2+2x-1}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{2x+1}{\sqrt{x^2+2x-1}}\,dx
\displaystyle =\int \frac{(2x+2)-1}{\sqrt{x^2+2x-1}}\,dx
\displaystyle =\int \frac{2x+2}{\sqrt{x^2+2x-1}}\,dx-\int \frac{dx}{\sqrt{x^2+2x-1}}
\displaystyle =\int \frac{2x+2}{\sqrt{x^2+2x-1}}\,dx-\int \frac{dx}{\sqrt{x^2+2x+1-1-1}}
\displaystyle =\int \frac{2x+2}{\sqrt{x^2+2x-1}}\,dx-\int \frac{dx}{\sqrt{(x+1)^2-(\sqrt2)^2}}
\displaystyle \text{Let } x^2+2x-1=t
\displaystyle \Rightarrow (2x+2)\,dx=dt
\displaystyle I=\int \frac{dt}{\sqrt{t}}-\int \frac{dx}{\sqrt{(x+1)^2-(\sqrt2)^2}}
\displaystyle =2\sqrt{t}-\log\left|x+1+\sqrt{(x+1)^2-(\sqrt2)^2}\right|+C
\displaystyle =2\sqrt{x^2+2x-1}-\log\left|x+1+\sqrt{x^2+2x-1}\right|+C

\displaystyle \textbf{Question 3: }~\int \frac{x+1}{\sqrt{4+5x-x^2}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{x+1}{\sqrt{4+5x-x^2}}\,dx
\displaystyle \text{Also } x+1=A\frac{d}{dx}(4+5x-x^2)+B
\displaystyle x+1=A(5-2x)+B
\displaystyle x+1=(-2A)x+5A+B
\displaystyle \text{Equating coefficients of like terms}
\displaystyle -2A=1
\displaystyle A=-\frac12
\displaystyle 5A+B=1
\displaystyle -\frac52+B=1
\displaystyle B=\frac72
\displaystyle I=\int \frac{-\frac12(5-2x)+\frac72}{\sqrt{4+5x-x^2}}\,dx
\displaystyle =-\frac12\int \frac{5-2x}{\sqrt{4+5x-x^2}}\,dx+\frac72\int \frac{dx}{\sqrt{4+5x-x^2}}
\displaystyle =-\frac12\int \frac{5-2x}{\sqrt{4+5x-x^2}}\,dx+\frac72\int \frac{dx}{\sqrt{4-(x^2-5x)}}
\displaystyle =-\frac12\int \frac{5-2x}{\sqrt{4+5x-x^2}}\,dx+\frac72\int \frac{dx}{\sqrt{4-\left(x-\frac52\right)^2+\frac{25}{4}}}
\displaystyle =-\frac12\int \frac{5-2x}{\sqrt{4+5x-x^2}}\,dx+\frac72\int \frac{dx}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^2-\left(x-\frac52\right)^2}}
\displaystyle \text{Let } 4+5x-x^2=t
\displaystyle \Rightarrow (5-2x)\,dx=dt
\displaystyle I=-\frac12\int \frac{dt}{\sqrt{t}}+\frac72\int \frac{dx}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^2-\left(x-\frac52\right)^2}}
\displaystyle =-\sqrt{t}+\frac72\sin^{-1}\left(\frac{x-\frac52}{\frac{\sqrt{41}}{2}}\right)+C
\displaystyle =-\sqrt{4+5x-x^2}+\frac72\sin^{-1}\left(\frac{2x-5}{\sqrt{41}}\right)+C

\displaystyle \textbf{Question 4: }~\int \frac{6x-5}{\sqrt{3x^2-5x+1}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{6x-5}{\sqrt{3x^2-5x+1}}\,dx
\displaystyle \text{Putting } 3x^2-5x+1=t
\displaystyle \Rightarrow (6x-5)\,dx=dt
\displaystyle \text{Then,}
\displaystyle I=\int \frac{dt}{\sqrt{t}}
\displaystyle =2\sqrt{t}+C
\displaystyle =2\sqrt{3x^2-5x+1}+C

\displaystyle \textbf{Question 5: }~\int \frac{3x+1}{\sqrt{5-2x-x^2}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{3x+1}{\sqrt{5-2x-x^2}}\,dx
\displaystyle \text{Consider, } 3x+1=A\frac{d}{dx}(5-2x-x^2)+B
\displaystyle \Rightarrow 3x+1=A(-2-2x)+B
\displaystyle \Rightarrow 3x+1=(-2A)x-2A+B
\displaystyle \text{Equating coefficients of like terms}
\displaystyle -2A=3
\displaystyle A=-\frac32
\displaystyle -2A+B=1
\displaystyle -2\left(-\frac32\right)+B=1
\displaystyle B=-2
\displaystyle \therefore I=\int \frac{-\frac32(-2-2x)-2}{\sqrt{5-2x-x^2}}\,dx
\displaystyle =-\frac32\int \frac{-2-2x}{\sqrt{5-2x-x^2}}\,dx-2\int \frac{dx}{\sqrt{5-2x-x^2}}
\displaystyle =-\frac32\int \frac{-2-2x}{\sqrt{5-2x-x^2}}\,dx-2\int \frac{dx}{\sqrt{5-(x^2+2x)}}
\displaystyle =-\frac32\int \frac{-2-2x}{\sqrt{5-\left[x^2+2x+1-1\right]}}\,dx-2\int \frac{dx}{\sqrt{6-(x+1)^2}}
\displaystyle =-\frac32\int \frac{-2-2x}{\sqrt{6-(x+1)^2}}\,dx-2\int \frac{dx}{\sqrt{(\sqrt6)^2-(x+1)^2}}
\displaystyle \text{Let } 5-2x-x^2=t
\displaystyle \Rightarrow (-2-2x)\,dx=dt
\displaystyle I=-\frac32\int \frac{dt}{\sqrt{t}}-2\int \frac{dx}{\sqrt{(\sqrt6)^2-(x+1)^2}}
\displaystyle =-\frac32\cdot2\sqrt{t}-2\sin^{-1}\left(\frac{x+1}{\sqrt6}\right)+C
\displaystyle =-3\sqrt{5-2x-x^2}-2\sin^{-1}\left(\frac{x+1}{\sqrt6}\right)+C

\displaystyle \textbf{Question 6: }~\int \frac{x}{\sqrt{8+x-x^2}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{x\,dx}{\sqrt{8+x-x^2}}
\displaystyle \text{Consider, } x=A\frac{d}{dx}(8+x-x^2)+B
\displaystyle x=A(1-2x)+B
\displaystyle x=(-2A)x+A+B
\displaystyle \text{Equating coefficients of like terms}
\displaystyle -2A=1
\displaystyle A=-\frac12
\displaystyle A+B=0
\displaystyle -\frac12+B=0
\displaystyle B=\frac12
\displaystyle \therefore x=-\frac12(1-2x)+\frac12
\displaystyle \text{Then,}
\displaystyle I=-\frac12\int \frac{1-2x}{\sqrt{8+x-x^2}}\,dx+\frac12\int \frac{dx}{\sqrt{8+x-x^2}}
\displaystyle =-\frac12\int \frac{1-2x}{\sqrt{8+x-x^2}}\,dx+\frac12\int \frac{dx}{\sqrt{8-(x^2-x)}}
\displaystyle =-\frac12\int \frac{1-2x}{\sqrt{8+x-x^2}}\,dx+\frac12\int \frac{dx}{\sqrt{8+\frac14-\left(x-\frac12\right)^2}}
\displaystyle =-\frac12\int \frac{1-2x}{\sqrt{8+x-x^2}}\,dx+\frac12\int \frac{dx}{\sqrt{\left(\frac{\sqrt{33}}{2}\right)^2-\left(x-\frac12\right)^2}}
\displaystyle \text{Let } 8+x-x^2=t
\displaystyle \Rightarrow (1-2x)\,dx=dt
\displaystyle I=-\frac12\int \frac{dt}{\sqrt{t}}+\frac12\int \frac{dx}{\sqrt{\left(\frac{\sqrt{33}}{2}\right)^2-\left(x-\frac12\right)^2}}
\displaystyle =-\frac12\cdot2\sqrt{t}+\frac12\sin^{-1}\left(\frac{x-\frac12}{\frac{\sqrt{33}}{2}}\right)+C
\displaystyle =-\sqrt{8+x-x^2}+\frac12\sin^{-1}\left(\frac{2x-1}{\sqrt{33}}\right)+C

\displaystyle \textbf{Question 7: }~\int \frac{x+2}{\sqrt{x^2+2x-1}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{(x+2)\,dx}{\sqrt{x^2+2x-1}}
\displaystyle \text{Consider,}
\displaystyle x+2 = A\,\frac{d}{dx}(x^2+2x-1) + B
\displaystyle \Rightarrow x+2 = A(2x+2) + B
\displaystyle \Rightarrow x+2 = (2A)x + 2A + B
\displaystyle \text{Equating coefficients of like terms,}
\displaystyle 2A = 1
\displaystyle \Rightarrow A = \frac{1}{2}
\displaystyle \text{And}
\displaystyle 2A + B = 2
\displaystyle \Rightarrow 2\left(\frac{1}{2}\right) + B = 2
\displaystyle \Rightarrow B = 1
\displaystyle \text{Then,}
\displaystyle I = \int \frac{\left[\frac{1}{2}(2x+2)+1\right]}{\sqrt{x^2+2x-1}}\,dx
\displaystyle = \frac{1}{2}\int \frac{(2x+2)\,dx}{\sqrt{x^2+2x-1}} + \int \frac{dx}{\sqrt{x^2+2x-1}}
\displaystyle \text{Let } x^2+2x-1 = t
\displaystyle \Rightarrow (2x+2)\,dx = dt
\displaystyle \therefore I = \frac{1}{2}\int \frac{dt}{\sqrt{t}} + \int \frac{dx}{\sqrt{x^2+2x-1}}
\displaystyle = \frac{1}{2}\int t^{-1/2}\,dt + \int \frac{dx}{\sqrt{(x+1)^2-(\sqrt{2})^2}}
\displaystyle = \frac{1}{2}\left(\frac{t^{1/2}}{\frac{1}{2}}\right) + \log\left|x+1+\sqrt{(x+1)^2-(\sqrt{2})^2}\right| + C
\displaystyle = \sqrt{t} + \log\left|x+1+\sqrt{x^2+2x-1}\right| + C
\displaystyle = \sqrt{x^2+2x-1} + \log\left|x+1+\sqrt{x^2+2x-1}\right| + C

\displaystyle \textbf{Question 8: }~\int \frac{x+2}{\sqrt{x^2-1}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{x+2}{\sqrt{x^2-1}}\,dx
\displaystyle = \int \frac{x}{\sqrt{x^2-1}}\,dx + 2\int \frac{dx}{\sqrt{x^2-1}}
\displaystyle \text{Let } x^2-1 = t
\displaystyle \Rightarrow 2x\,dx = dt
\displaystyle \Rightarrow x\,dx = \frac{dt}{2}
\displaystyle \text{Then,}
\displaystyle I = \frac{1}{2}\int \frac{dt}{\sqrt{t}} + 2\int \frac{dx}{\sqrt{x^2-1}}
\displaystyle = \frac{1}{2}\int t^{-1/2}\,dt + 2\int \frac{dx}{\sqrt{x^2-1}}
\displaystyle = \frac{1}{2}\left(\frac{t^{1/2}}{\frac{1}{2}}\right) + 2\log\left|x+\sqrt{x^2-1}\right| + C
\displaystyle = \sqrt{t} + 2\log\left|x+\sqrt{x^2-1}\right| + C
\displaystyle = \sqrt{x^2-1} + 2\log\left|x+\sqrt{x^2-1}\right| + C

\displaystyle \textbf{Question 9: }~\int \frac{x-1}{\sqrt{x^2+1}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{x-1}{\sqrt{x^2+1}}\,dx
\displaystyle = \int \frac{x\,dx}{\sqrt{x^2+1}} - \int \frac{dx}{\sqrt{x^2+1}}
\displaystyle \text{Putting } x^2+1 = t
\displaystyle \Rightarrow 2x\,dx = dt
\displaystyle \Rightarrow x\,dx = \frac{dt}{2}
\displaystyle \text{Then,}
\displaystyle I = \frac{1}{2}\int \frac{dt}{\sqrt{t}} - \int \frac{dx}{\sqrt{x^2+1}}
\displaystyle = \frac{1}{2}\int t^{-1/2}\,dt - \int \frac{dx}{\sqrt{x^2+1}}
\displaystyle = \frac{1}{2}\left(\frac{t^{1/2}}{\frac{1}{2}}\right) - \log\left|x+\sqrt{x^2+1}\right| + C
\displaystyle = \sqrt{t} - \log\left|x+\sqrt{x^2+1}\right| + C
\displaystyle = \sqrt{x^2+1} - \log\left|x+\sqrt{x^2+1}\right| + C

\displaystyle \textbf{Question 10: }~\int \frac{x}{\sqrt{x^2+x+1}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{x\,dx}{\sqrt{x^2+x+1}}
\displaystyle \text{Consider,}
\displaystyle x = A\frac{d}{dx}(x^2+x+1) + B
\displaystyle \Rightarrow x = A(2x+1) + B
\displaystyle \Rightarrow x = (2A)x + A + B
\displaystyle \text{Equating coefficients of like terms,}
\displaystyle 2A = 1
\displaystyle \Rightarrow A = \frac{1}{2}
\displaystyle \text{And}
\displaystyle A + B = 0
\displaystyle \Rightarrow \frac{1}{2} + B = 0
\displaystyle \Rightarrow B = -\frac{1}{2}
\displaystyle \therefore I = \int \frac{\left(\frac{1}{2}(2x+1)-\frac{1}{2}\right)}{\sqrt{x^2+x+1}}\,dx
\displaystyle = \frac{1}{2}\int \frac{2x+1}{\sqrt{x^2+x+1}}\,dx - \frac{1}{2}\int \frac{dx}{\sqrt{x^2+x+1}}
\displaystyle \text{Putting } x^2+x+1 = t
\displaystyle \Rightarrow (2x+1)\,dx = dt
\displaystyle \text{Then,}
\displaystyle I = \frac{1}{2}\int \frac{dt}{\sqrt{t}} - \frac{1}{2}\int \frac{dx}{\sqrt{x^2+x+1}}
\displaystyle = \frac{1}{2}\int t^{-1/2}\,dt - \frac{1}{2}\int \frac{dx}{\sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}}
\displaystyle = \frac{1}{2}\left(\frac{t^{1/2}}{\frac{1}{2}}\right) - \frac{1}{2}\log\left|x+\frac{1}{2}+\sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\right| + C
\displaystyle = \sqrt{t} - \frac{1}{2}\log\left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right| + C
\displaystyle = \sqrt{x^2+x+1} - \frac{1}{2}\log\left|x+\frac{1}{2}+\sqrt{x^2+x+1}\right| + C

\displaystyle \textbf{Question 11: }~\int \frac{x+1}{\sqrt{x^2+1}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{x+1}{\sqrt{x^2+1}}\,dx
\displaystyle = \int \frac{x\,dx}{\sqrt{x^2+1}} + \int \frac{dx}{\sqrt{x^2+1}}
\displaystyle \text{Putting } x^2+1 = t
\displaystyle \Rightarrow 2x\,dx = dt
\displaystyle \Rightarrow x\,dx = \frac{dt}{2}
\displaystyle \text{Then,}
\displaystyle I = \frac{1}{2}\int \frac{dt}{\sqrt{t}} + \int \frac{dx}{\sqrt{x^2+1}}
\displaystyle = \frac{1}{2}\int t^{-1/2}\,dt + \int \frac{dx}{\sqrt{x^2+1}}
\displaystyle = \frac{1}{2}\left(\frac{t^{1/2}}{\frac{1}{2}}\right) + \log\left|x+\sqrt{x^2+1}\right| + C
\displaystyle = \sqrt{t} + \log\left|x+\sqrt{x^2+1}\right| + C
\displaystyle = \sqrt{x^2+1} + \log\left|x+\sqrt{x^2+1}\right| + C

\displaystyle \textbf{Question 12: }~\int \frac{2x+5}{\sqrt{x^2+2x+5}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{2x+5}{\sqrt{x^2+2x+5}}\,dx
\displaystyle \text{Consider,}
\displaystyle 2x+5 = A\frac{d}{dx}(x^2+2x+5) + B
\displaystyle \Rightarrow 2x+5 = A(2x+2) + B
\displaystyle \Rightarrow 2x+5 = (2A)x + 2A + B
\displaystyle \text{Equating coefficients of like terms,}
\displaystyle 2A = 2 \Rightarrow A = 1
\displaystyle \text{And}
\displaystyle 2A + B = 5 \Rightarrow B = 3
\displaystyle \therefore I = \int \frac{(2x+2)+3}{\sqrt{x^2+2x+5}}\,dx
\displaystyle = \int \frac{(2x+2)\,dx}{\sqrt{x^2+2x+5}} + 3\int \frac{dx}{\sqrt{x^2+2x+5}}
\displaystyle \text{Let } x^2+2x+5 = t
\displaystyle \Rightarrow (2x+2)\,dx = dt
\displaystyle \text{Then,}
\displaystyle I = \int \frac{dt}{\sqrt{t}} + 3\int \frac{dx}{\sqrt{x^2+2x+5}}
\displaystyle = \int t^{-1/2}\,dt + 3\int \frac{dx}{\sqrt{(x+1)^2+2^2}}
\displaystyle = \frac{t^{1/2}}{\frac{1}{2}} + 3\log\left|x+1+\sqrt{(x+1)^2+2^2}\right| + C
\displaystyle = 2\sqrt{t} + 3\log\left|x+1+\sqrt{x^2+2x+5}\right| + C
\displaystyle = 2\sqrt{x^2+2x+5} + 3\log\left|x+1+\sqrt{x^2+2x+5}\right| + C

\displaystyle \textbf{Question 13: }~\int \frac{3x+1}{\sqrt{5-2x-x^2}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{3x+1}{\sqrt{5-2x-x^2}}\,dx
\displaystyle \text{Consider,}
\displaystyle 3x+1 = A\frac{d}{dx}(5-2x-x^2) + B
\displaystyle \Rightarrow 3x+1 = A(-2-2x) + B
\displaystyle \Rightarrow 3x+1 = (-2A)x + (-2A + B)
\displaystyle \text{Equating coefficients of like terms,}
\displaystyle -2A = 3
\displaystyle \Rightarrow A = -\frac{3}{2}
\displaystyle \text{And}
\displaystyle -2A + B = 1
\displaystyle \Rightarrow -2\left(-\frac{3}{2}\right) + B = 1
\displaystyle \Rightarrow B = -2
\displaystyle \therefore I = \int \frac{\left[-\frac{3}{2}(-2-2x)-2\right]}{\sqrt{5-2x-x^2}}\,dx
\displaystyle = -\frac{3}{2}\int \frac{(-2-2x)\,dx}{\sqrt{5-2x-x^2}} - 2\int \frac{dx}{\sqrt{5-2x-x^2}}
\displaystyle = -\frac{3}{2}\int \frac{(-2-2x)\,dx}{\sqrt{5-(x^2+2x)}} - 2\int \frac{dx}{\sqrt{5-(x^2+2x)}}
\displaystyle = -\frac{3}{2}\int \frac{(-2-2x)\,dx}{\sqrt{5-(x^2+2x+1-1)}} - 2\int \frac{dx}{\sqrt{5-(x+1)^2}}
\displaystyle = -\frac{3}{2}\int \frac{(-2-2x)\,dx}{\sqrt{5-2x-x^2}} - 2\int \frac{dx}{\sqrt{(\sqrt{6})^2-(x+1)^2}}
\displaystyle \text{Putting } 5-2x-x^2 = t
\displaystyle \Rightarrow (-2-2x)\,dx = dt
\displaystyle \text{Then,}
\displaystyle I = -\frac{3}{2}\int \frac{dt}{\sqrt{t}} - 2\sin^{-1}\left(\frac{x+1}{\sqrt{6}}\right) + C
\displaystyle = -\frac{3}{2}\cdot 2\sqrt{t} - 2\sin^{-1}\left(\frac{x+1}{\sqrt{6}}\right) + C
\displaystyle = -3\sqrt{5-2x-x^2} - 2\sin^{-1}\left(\frac{x+1}{\sqrt{6}}\right) + C

\displaystyle \textbf{Question 14: }~\int \sqrt{\frac{1-x}{1+x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \sqrt{\frac{1-x}{1+x}}\,dx
\displaystyle = \int \sqrt{\frac{(1-x)(1-x)}{(1+x)(1-x)}}\,dx
\displaystyle = \int \frac{1-x}{\sqrt{1-x^2}}\,dx
\displaystyle = \int \frac{dx}{\sqrt{1-x^2}} - \int \frac{x\,dx}{\sqrt{1-x^2}}
\displaystyle \text{Putting } 1-x^2 = t
\displaystyle \Rightarrow -2x\,dx = dt
\displaystyle \Rightarrow x\,dx = -\frac{dt}{2}
\displaystyle \text{Then,}
\displaystyle I = \int \frac{dx}{\sqrt{1-x^2}} + \frac{1}{2}\int \frac{dt}{\sqrt{t}}
\displaystyle = \sin^{-1}(x) + \frac{1}{2}\cdot 2\sqrt{t} + C
\displaystyle = \sin^{-1}(x) + \sqrt{t} + C
\displaystyle = \sin^{-1}(x) + \sqrt{1-x^2} + C

\displaystyle \textbf{Question 15: }~\int \frac{2x+1}{\sqrt{x^2+4x+3}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{2x+1}{\sqrt{x^2+4x+3}}\,dx
\displaystyle \text{Consider,}
\displaystyle 2x+1 = A\frac{d}{dx}(x^2+4x+3) + B
\displaystyle \Rightarrow 2x+1 = A(2x+4) + B
\displaystyle \Rightarrow 2x+1 = (2A)x + 4A + B
\displaystyle \text{Equating coefficients of like terms,}
\displaystyle 2A = 2
\displaystyle \Rightarrow A = 1
\displaystyle \text{And}
\displaystyle 4A + B = 1
\displaystyle \Rightarrow 4 + B = 1
\displaystyle \Rightarrow B = -3
\displaystyle \therefore I = \int \frac{(2x+4)-3}{\sqrt{x^2+4x+3}}\,dx
\displaystyle = \int \frac{(2x+4)\,dx}{\sqrt{x^2+4x+3}} - 3\int \frac{dx}{\sqrt{x^2+4x+3}}
\displaystyle \text{Let } x^2+4x+3 = t
\displaystyle \Rightarrow (2x+4)\,dx = dt
\displaystyle \text{Then,}
\displaystyle I = \int \frac{dt}{\sqrt{t}} - 3\int \frac{dx}{\sqrt{(x+2)^2-1^2}}
\displaystyle = \int t^{-1/2}\,dt - 3\int \frac{dx}{\sqrt{(x+2)^2-1^2}}
\displaystyle = \frac{t^{1/2}}{\frac{1}{2}} - 3\log\left|x+2+\sqrt{(x+2)^2-1}\right| + C
\displaystyle = 2\sqrt{t} - 3\log\left|x+2+\sqrt{x^2+4x+3}\right| + C
\displaystyle = 2\sqrt{x^2+4x+3} - 3\log\left|x+2+\sqrt{x^2+4x+3}\right| + C

\displaystyle \textbf{Question 16: }~\int \frac{2x+3}{\sqrt{x^2+4x+5}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{2x+3}{\sqrt{x^2+4x+5}}\,dx
\displaystyle = \int \frac{(2x+4)-1}{\sqrt{x^2+4x+5}}\,dx
\displaystyle = \int \frac{(2x+4)\,dx}{\sqrt{x^2+4x+5}} - \int \frac{dx}{\sqrt{x^2+4x+5}}
\displaystyle = \int \frac{(2x+4)\,dx}{\sqrt{x^2+4x+5}} - \int \frac{dx}{\sqrt{(x+2)^2+1}}
\displaystyle \text{Consider,}
\displaystyle x^2+4x+5 = t
\displaystyle \Rightarrow (2x+4)\,dx = dt
\displaystyle \therefore I = \int \frac{dt}{\sqrt{t}} - \int \frac{dx}{\sqrt{(x+2)^2+1}}
\displaystyle = \int t^{-1/2}\,dt - \int \frac{dx}{\sqrt{(x+2)^2+1^2}}
\displaystyle = \frac{t^{1/2}}{\frac{1}{2}} - \log\left|x+2+\sqrt{(x+2)^2+1}\right| + C
\displaystyle = 2\sqrt{t} - \log\left|x+2+\sqrt{x^2+4x+5}\right| + C
\displaystyle = 2\sqrt{x^2+4x+5} - \log\left|x+2+\sqrt{x^2+4x+5}\right| + C

\displaystyle \textbf{Question 17: }~\int \frac{5x+3}{\sqrt{x^2+4x+10}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{5x+3}{\sqrt{x^2+4x+10}}\,dx
\displaystyle \text{Consider,}
\displaystyle 5x+3 = A\frac{d}{dx}(x^2+4x+10) + B
\displaystyle \Rightarrow 5x+3 = A(2x+4) + B
\displaystyle \Rightarrow 5x+3 = (2A)x + 4A + B
\displaystyle \text{Equating coefficients of like terms,}
\displaystyle 2A = 5
\displaystyle \Rightarrow A = \frac{5}{2}
\displaystyle \text{And}
\displaystyle 4A + B = 3
\displaystyle \Rightarrow 4\times\frac{5}{2} + B = 3
\displaystyle \Rightarrow B = -7
\displaystyle \therefore I = \int \frac{\left(\frac{5}{2}(2x+4)-7\right)}{\sqrt{x^2+4x+10}}\,dx
\displaystyle = \frac{5}{2}\int \frac{(2x+4)\,dx}{\sqrt{x^2+4x+10}} - 7\int \frac{dx}{\sqrt{x^2+4x+10}}
\displaystyle = \frac{5}{2}\int \frac{(2x+4)\,dx}{\sqrt{x^2+4x+10}} - 7\int \frac{dx}{\sqrt{(x+2)^2+(\sqrt{6})^2}}
\displaystyle \text{Putting } x^2+4x+10 = t
\displaystyle \Rightarrow (2x+4)\,dx = dt
\displaystyle \text{Then,}
\displaystyle I = \frac{5}{2}\int \frac{dt}{\sqrt{t}} - 7\log\left|x+2+\sqrt{(x+2)^2+6}\right| + C
\displaystyle = \frac{5}{2}\int t^{-1/2}\,dt - 7\log\left|x+2+\sqrt{x^2+4x+10}\right| + C
\displaystyle = \frac{5}{2}\cdot 2\sqrt{t} - 7\log\left|x+2+\sqrt{x^2+4x+10}\right| + C
\displaystyle = 5\sqrt{x^2+4x+10} - 7\log\left|x+2+\sqrt{x^2+4x+10}\right| + C

\displaystyle \textbf{Question 18: }~\int \frac{x+2}{\sqrt{x^2+2x+3}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I = \int \frac{x+2}{\sqrt{x^2+2x+3}}\,dx
\displaystyle \text{We express } x+2 = A\frac{d}{dx}(x^2+2x+3) + B
\displaystyle \Rightarrow x+2 = A(2x+2) + B
\displaystyle \text{Equating coefficients of } x \text{ and constants, we get}
\displaystyle 2A = 1 \text{ and } 2A + B = 2
\displaystyle \Rightarrow A = \frac{1}{2} \text{ and } B = 1
\displaystyle \therefore I = \int \frac{\frac{1}{2}(2x+2)+1}{\sqrt{x^2+2x+3}}\,dx
\displaystyle = \frac{1}{2}\int \frac{2x+2}{\sqrt{x^2+2x+3}}\,dx + \int \frac{dx}{\sqrt{x^2+2x+3}}
\displaystyle \text{Let } I_1 = \int \frac{2x+2}{\sqrt{x^2+2x+3}}\,dx
\displaystyle \text{Let } x^2+2x+3 = u
\displaystyle \Rightarrow (2x+2)\,dx = du
\displaystyle \therefore I_1 = \int \frac{du}{\sqrt{u}}
\displaystyle = 2\sqrt{u} + c_1
\displaystyle = 2\sqrt{x^2+2x+3} + c_1
\displaystyle \text{And let } I_2 = \int \frac{dx}{\sqrt{x^2+2x+3}}
\displaystyle = \int \frac{dx}{\sqrt{(x+1)^2+(\sqrt{2})^2}}
\displaystyle \text{Let } (x+1) = u
\displaystyle \Rightarrow dx = du
\displaystyle \therefore I_2 = \int \frac{du}{\sqrt{u^2+(\sqrt{2})^2}}
\displaystyle = \log\left|u+\sqrt{u^2+(\sqrt{2})^2}\right| + c_2
\displaystyle = \log\left|(x+1)+\sqrt{x^2+2x+3}\right| + c_2
\displaystyle \text{From } I = \frac{1}{2}I_1 + I_2, \text{ we get}
\displaystyle I = \frac{1}{2}\left(2\sqrt{x^2+2x+3}+c_1\right) + \log\left|(x+1)+\sqrt{x^2+2x+3}\right| + c_2
\displaystyle = \sqrt{x^2+2x+3} + \log\left|(x+1)+\sqrt{x^2+2x+3}\right| + C

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