\displaystyle \textbf{Question 1: } \text{If } \int \sin^{-1}(\cos x)\,dx = g(x) - \frac{x^{2}}{2} + C, \text{ then } g(x) \text{ will be }   
\displaystyle \text{(a) } \frac{\pi}{2}x \quad \text{(b) } -\frac{\pi}{2}x \quad \text{(c) } 0 \quad \text{(d) } \frac{x}{2}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given, } \int \sin^{-1}(\cos x)\,dx=g(x)-\frac{x^{2}}{2}+C \quad (i)
\displaystyle \sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}
\displaystyle \Rightarrow \sin^{-1}(\cos x)+\cos^{-1}(\cos x)=\frac{\pi}{2}
\displaystyle \Rightarrow \sin^{-1}(\cos x)=\frac{\pi}{2}-\cos^{-1}(\cos x)=\frac{\pi}{2}-x
\displaystyle \therefore \int \sin^{-1}(\cos x)\,dx=\int \left(\frac{\pi}{2}-x\right)dx=\frac{\pi x}{2}-\frac{x^{2}}{2}+C \quad (ii)
\displaystyle \text{From Eqs. (i) and (ii), we get } g(x)=\frac{\pi x}{2}
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\displaystyle \textbf{Question 2: } \text{If } \int \log(2x)\,dx = x\log(2x) - k + C, \text{ where } k \text{ is a function of } x, \text{ then find } k
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \log 2x\,dx
\displaystyle \text{On putting } 2x=k \Rightarrow 2dx=dk \Rightarrow dx=\frac{dk}{2}
\displaystyle \therefore \int \log 2x\,dx=\int \log k\cdot \frac{dk}{2}=\frac{1}{2}\int \log k\,dk
\displaystyle =\frac{1}{2}[k\log k-k]+C
\displaystyle =\frac{1}{2}[2x\log(2x)-2x]+C
\displaystyle \Rightarrow \int \log 2x\,dx=x\log 2x-x+C
\displaystyle \text{On comparing with } \int \log 2x\,dx=x\log 2x-k+C,\text{ we get } k=x
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\displaystyle \textbf{Question 3: } \int a^{3x+2}\,dx \text{ is equal to }   
\displaystyle \text{(a) } \frac{e^{3x}}{3\log_{e}a}+C \quad \text{(b) } a^{2x} + \left(\frac{a^{3x}}{3\log_{e}a}\right)+C \quad \text{(c) } a^{2}\left(\frac{a^{3x}}{3\log_{e}a}\right)+C \quad \text{(d) } a^{2}\left(\frac{a^{3x}}{\log_{e}a}\right)+C
\displaystyle \text{Answer:}
\displaystyle \text{(c) Let } I=\int a^{3x+2}\,dx=\int a^{3x}a^{2}\,dx=a^{2}\int a^{3x}\,dx
\displaystyle =a^{2}\cdot \frac{a^{3x}}{3\log a}+C
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\displaystyle \textbf{Question 4: } \int e^{\sin x}\cos x\,dx \text{ is equal to }   
\displaystyle \text{(a) } e^{\cos x}+C \quad \text{(b) } e^{\sin x}+C \quad \text{(c) } \frac{\sin^{2}x}{2}+C \quad \text{(d) } e^{\sin^{2}x}+C
\displaystyle \text{Answer:}
\displaystyle \text{(b) Let } I=\int e^{\sin x}\cos x\,dx
\displaystyle \text{Put } \sin x=t,\text{ then } \cos x\,dx=dt
\displaystyle \Rightarrow I=\int e^{t}\,dt=e^{t}+C=e^{\sin x}+C
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\displaystyle \textbf{Question 5: } \int \frac{\sin 2x}{\cos x}\,dx \text{ is equal to } \quad   
\displaystyle \text{(a) } -2\cos x+C \quad \text{(b) } 2\cos x+C \quad \text{(c) } \frac{-\cos x}{2}+C \quad \text{(d) } \frac{\cos x}{2}+C
\displaystyle \text{Answer:}
\displaystyle \text{(a) Let } I=\int \frac{\sin 2x}{\cos x}\,dx=\int \frac{2\sin x\cos x}{\cos x}\,dx
\displaystyle =\int 2\sin x\,dx=2\int \sin x\,dx=-2\cos x+C
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\displaystyle \textbf{Question 6: } \text{Evaluate } \int \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{2}\,dx.  
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{2}dx
\displaystyle =\int \left(x+\frac{1}{x}+2\right)dx
\displaystyle =\int x\,dx+\int \frac{1}{x}\,dx+2\int dx
\displaystyle =\frac{x^{2}}{2}+\log|x|+2x+C
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\displaystyle \textbf{Question 7: } \text{Evaluate } \int \frac{1}{\sin^{2}x\cos^{2}x}\,dx.  
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{1}{\sin^{2}x\cos^{2}x}\,dx=\int \frac{\sin^{2}x+\cos^{2}x}{\sin^{2}x\cos^{2}x}\,dx
\displaystyle \text{[}\because \sin^{2}\theta+\cos^{2}\theta=1\text{]}
\displaystyle =\int \left(\frac{1}{\cos^{2}x}+\frac{1}{\sin^{2}x}\right)dx
\displaystyle =\int (\sec^{2}x+\mathrm{cosec}^{2}x)\,dx=\tan x-\cot x+C
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\displaystyle \textbf{Question 8: } \text{If } \int \frac{(\log x)^{2}}{x}\,dx = \frac{(\log x)^{k}}{k}+C, \text{ then the value of } k \text{ is }
\displaystyle \text{(a) } 3 \quad \text{(b) } 2 \quad \text{(c) } 1 \quad \text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Let } I=\int \frac{(\log x)^{2}}{x}\,dx
\displaystyle \text{On putting } \log x=t \Rightarrow \frac{1}{x}dx=dt
\displaystyle \therefore I=\int t^{2}\,dt=\frac{t^{3}}{3}+C
\displaystyle \Rightarrow I=\frac{(\log x)^{3}}{3}+C
\displaystyle \therefore \int \frac{(\log x)^{2}}{x}\,dx=\frac{(\log x)^{3}}{3}+C \Rightarrow k=3
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\displaystyle \textbf{Question 9: } \text{Given } \int e^{x} \left(\frac{x-1}{x^{2}}\right)\,dx=e^{x}f(x)+C. \text{ Then, } f(x) \text{ satisfying the equation is }  
\displaystyle \text{(a) } x \quad \text{(b) } x^{2} \quad \text{(c) } \frac{1}{x} \quad \text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \text{(c) Given, } \int e^{x}\left(\frac{x-1}{x^{2}}\right)dx=e^{x}f(x)+C \quad (i)
\displaystyle \text{Let } I=\int e^{x}\left(\frac{x-1}{x^{2}}\right)dx=\int e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right)dx
\displaystyle =\int e^{x}\left(\frac{1}{x}+\left(-\frac{1}{x^{2}}\right)\right)dx
\displaystyle =e^{x}\left(\frac{1}{x}\right)+C
\displaystyle \text{[}\because e^{x}(f(x)+f'(x))=e^{x}f(x)\text{]}
\displaystyle \therefore \int e^{x}\left(\frac{x-1}{x^{2}}\right)dx=e^{x}\left(\frac{1}{x}\right)+C \quad (ii)
\displaystyle \text{On comparing Eqs. (i) and (ii), we get } f(x)=\frac{1}{x}
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\displaystyle \textbf{Question 10: } \text{Evaluate } \int \frac{5-3x^{2}}{6-x^{3}+5x}\,dx. \quad   
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{5-3x^{2}}{6-x^{3}+5x}\,dx
\displaystyle \text{Let } 6-x^{3}+5x=t \Rightarrow (-3x^{2}+5)dx=dt
\displaystyle \therefore I=\int \frac{dt}{t}=\log|t|+C
\displaystyle \Rightarrow I=\log|6-x^{3}+5x|+C
\\

\displaystyle \textbf{Question 11: } \text{Evaluate } \int xe^{x}\,dx.  
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int xe^{x}\,dx=x\int e^{x}dx-\int \left(\frac{d}{dx}x\right)\left[\int e^{x}dx\right]dx
\displaystyle =xe^{x}-\int (1\cdot e^{x})dx=xe^{x}-e^{x}+C
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\displaystyle \textbf{Question 12: } \text{Evaluate } \int [\sin(\log x)+\cos(\log x)]\,dx. \quad   
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int [\sin(\log x)+\cos(\log x)]dx
\displaystyle =\int \sin(\log x)\,dx+\int \cos(\log x)\,dx
\displaystyle \text{Integrating by parts: } \int \sin(\log x)\,dx=x\sin(\log x)-\int x\cdot \frac{\cos(\log x)}{x}dx
\displaystyle =x\sin(\log x)-\int \cos(\log x)\,dx
\displaystyle \therefore I=x\sin(\log x)-\int \cos(\log x)\,dx+\int \cos(\log x)\,dx
\displaystyle =x\sin(\log x)+C
\\

\displaystyle \textbf{Question 13: } \text{Evaluate } \int \frac{x^{3}-x^{2}+x-1}{x-1}\,dx. \quad   
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{x^{3}-x^{2}+x-1}{x-1}\,dx
\displaystyle =\int \frac{x^{2}(x-1)+1(x-1)}{x-1}\,dx
\displaystyle =\int (x^{2}+1)\,dx
\displaystyle =\int x^{2}\,dx+\int 1\,dx
\displaystyle =\frac{x^{3}}{3}+x+C
\\

\displaystyle \textbf{Question 14: } \text{Evaluate } \int \log_{10}x\,dx.  
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \log_{10}x\,dx=\int \frac{\log x}{\log 10}\,dx
\displaystyle =\frac{1}{\log 10}\int \log x\,dx
\displaystyle =\frac{1}{\log 10}\left[x\log x-\int x\cdot \frac{1}{x}dx\right]+C
\displaystyle =\frac{1}{\log 10}(x\log x-x)+C
\\

\displaystyle \textbf{Question 15: } \text{Evaluate } \int \frac{\sec^{2}x}{\mathrm{cosec}\,x}\,dx. \quad 
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{\sec^{2}x}{\mathrm{cosec}^{2}x}\,dx=\int \frac{\sin^{2}x}{\cos^{2}x}\,dx
\displaystyle =\int \tan^{2}x\,dx=\int (\sec^{2}x-1)dx
\displaystyle =\int \sec^{2}x\,dx-\int dx=\tan x-x+C
\\

\displaystyle \textbf{Question 16: } \text{Evaluate } \int \sqrt{\mathrm{cosec}\,x-1}\,dx.   
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \sqrt{\mathrm{cosec}x-1}\,dx
\displaystyle =\int \sqrt{\frac{1}{\sin x}-1}\,dx
\displaystyle =\int \sqrt{\frac{1-\sin x}{\sin x}}\,dx
\\

\displaystyle \textbf{Question 17: } \text{Evaluate } \int x\tan^{-1}x\,dx.  
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int x\tan^{-1}x\,dx
\displaystyle \text{Integrating by parts, } I=x\left(\frac{x^{2}}{2}\right)-\int \frac{1}{1+x^{2}}\cdot \frac{x^{2}}{2}dx
\displaystyle =\frac{x^{2}}{2}\tan^{-1}x-\frac{1}{2}\int \frac{x^{2}}{1+x^{2}}dx
\displaystyle =\frac{x^{2}}{2}\tan^{-1}x-\frac{1}{2}\int \frac{x^{2}+1-1}{1+x^{2}}dx
\displaystyle =\frac{x^{2}}{2}\tan^{-1}x-\frac{1}{2}\int dx+\frac{1}{2}\int \frac{1}{1+x^{2}}dx
\displaystyle =\frac{x^{2}}{2}\tan^{-1}x-\frac{x}{2}+\frac{1}{2}\tan^{-1}x+C
\\

\displaystyle \textbf{Question 18: } \text{Evaluate } \int \frac{dx}{x\{(\log x)^{2}+5\log x+6\}}. \quad  
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{dx}{x[(\log x)^{2}+5\log x+6]}
\displaystyle \text{On putting } \log x=t \Rightarrow \frac{1}{x}dx=dt
\displaystyle \therefore I=\int \frac{dt}{t^{2}+5t+6}=\int \frac{dt}{(t+2)(t+3)}
\displaystyle \text{Let } \frac{1}{(t+2)(t+3)}=\frac{A}{t+2}+\frac{B}{t+3}
\displaystyle \Rightarrow 1=A(t+3)+B(t+2)
\displaystyle \text{On putting } t=-3,\ B=-1;\quad \text{on putting } t=-2,\ A=1
\displaystyle \therefore I=\int \left(\frac{1}{t+2}-\frac{1}{t+3}\right)dt
\displaystyle =\log|t+2|-\log|t+3|+C
\displaystyle =\log\left|\frac{t+2}{t+3}\right|+C
\displaystyle =\log\left|\frac{\log x+2}{\log x+3}\right|+C
\\

\displaystyle \textbf{Question 19: } \text{Evaluate } \int \sqrt{\frac{1+\sin x}{1-\sin x}}\,dx. \quad  
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{\sqrt{\cos^{2}\frac{x}{2}+\sin^{2}\frac{x}{2}-2\sin\frac{x}{2}\cos\frac{x}{2}}}{\sqrt{\sin x}}dx
\displaystyle =\int \frac{\sqrt{\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)^{2}}}{\sqrt{\sin x}}dx
\displaystyle =\int \frac{\cos\frac{x}{2}-\sin\frac{x}{2}}{\sqrt{\sin x}}dx
\displaystyle =\int \frac{1}{\sqrt{2}}\frac{\sqrt{2}\cos\frac{x}{2}-\sqrt{2}\sin\frac{x}{2}}{\sqrt{\sin x}}dx
\displaystyle =\sqrt{2}\int \frac{\cos\frac{x}{2}\cos\frac{\pi}{4}-\sin\frac{x}{2}\sin\frac{\pi}{4}}{\sqrt{\sin x}}dx
\displaystyle =\sqrt{2}\int \frac{\cos\left(\frac{\pi}{4}+\frac{x}{2}\right)}{\sqrt{\sin x}}dx
\displaystyle =\sqrt{2}\int \frac{\sin\left(\frac{\pi}{2}-\left(\frac{\pi}{4}+\frac{x}{2}\right)\right)}{\sqrt{\sin x}}dx
\displaystyle =\sqrt{2}\int \frac{\sin\left(\frac{\pi}{4}-\frac{x}{2}\right)}{\sqrt{\sin x}}dx \quad (i)
\displaystyle \text{Let } \sqrt{2}\cos\left(\frac{\pi}{4}-\frac{x}{2}\right)=t
\displaystyle \Rightarrow -\sqrt{2}\sin\left(\frac{\pi}{4}-\frac{x}{2}\right)\left(-\frac{1}{2}\right)dx=dt
\displaystyle \Rightarrow \sin\left(\frac{\pi}{4}-\frac{x}{2}\right)dx=\sqrt{2}\,dt
\displaystyle \text{Substituting in (i), } I=\int \frac{\sqrt{2}\,dt}{\sqrt{\sin x}}
\displaystyle \text{Using } \sin x=2\sin\frac{x}{2}\cos\frac{x}{2}
\displaystyle \Rightarrow I=\int \frac{\sqrt{2}\,dt}{\sqrt{2\sin\frac{x}{2}\cos\frac{x}{2}}}=\int \frac{dt}{\sqrt{\sin\frac{x}{2}\cos\frac{x}{2}}}
\displaystyle \text{Simplifying gives } I=2\log\left|\sqrt{2}\cos\left(\frac{\pi}{4}-\frac{x}{2}\right)+\sqrt{2\cos^{2}\left(\frac{\pi}{4}-\frac{x}{2}\right)-1}\right|+C
\displaystyle =2\log\left|\cos\frac{x}{2}+\sin\frac{x}{2}+\sqrt{\sin x}\right|+C
\displaystyle \text{Let } I=\int \sqrt{\frac{1+\sin x}{1-\sin x}}\,dx
\displaystyle =\int \sqrt{\frac{(1+\sin x)^{2}}{1-\sin^{2}x}}\,dx
\displaystyle =\int \sqrt{\frac{(1+\sin x)^{2}}{\cos^{2}x}}\,dx
\displaystyle =\int \frac{1+\sin x}{\cos x}\,dx
\displaystyle =\int (\sec x+\tan x)\,dx
\displaystyle =\log|\sec x+\tan x|-\log|\cos x|+C
\\

\displaystyle \textbf{Question 20: } \text{Evaluate } \int \frac{6x+7}{\sqrt{(x-5)(x-4)}}\,dx.   
\displaystyle \text{Answer:}
\displaystyle \text{We have, } I=\int \frac{6x+7}{\sqrt{(x-5)(x-4)}}\,dx
\displaystyle \Rightarrow I=\int \frac{6x+7}{\sqrt{x^{2}-9x+20}}\,dx
\displaystyle \text{Let } 6x+7=A\frac{d}{dx}(x^{2}-9x+20)+B
\displaystyle \Rightarrow 6x+7=A(2x-9)+B
\displaystyle \Rightarrow 6x+7=2Ax-9A+B
\displaystyle \text{Equating coefficients of }x\text{ and constant terms, we get } A=3,\ B=34
\displaystyle \therefore I=\int \frac{3(2x-9)}{\sqrt{x^{2}-9x+20}}\,dx+\int \frac{34}{\sqrt{x^{2}-9x+20}}\,dx
\displaystyle \text{On putting } x^{2}-9x+20=t \Rightarrow (2x-9)dx=dt
\displaystyle I=\int \frac{3dt}{\sqrt{t}}+34\log\left|x-\frac{9}{2}+\sqrt{\left(x-\frac{9}{2}\right)^{2}-\frac{1}{4}}\right|+C
\displaystyle =6\sqrt{t}+34\log\left|x-\frac{9}{2}+\sqrt{x^{2}-9x+20}\right|+C
\displaystyle \Rightarrow I=6\sqrt{x^{2}-9x+20}+34\log\left|x-\frac{9}{2}+\sqrt{x^{2}-9x+20}\right|+C
\\

\displaystyle \textbf{Question 21: } \text{Evaluate } \int \frac{x^{2}}{x^{4}+1}\,dx.   
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{x^{2}}{1+x^{4}}\,dx
\displaystyle =\int \frac{1}{x^{2}+\frac{1}{x^{2}}}\,dx
\displaystyle =\frac{1}{2}\int \left[\frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}}+\frac{1-\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}}\right]dx
\displaystyle =\frac{1}{2}\int \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}}dx+\frac{1}{2}\int \frac{1-\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}}dx
\displaystyle =I_{1}+I_{2}
\displaystyle I_{1}=\frac{1}{2}\int \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}}\,dx
\displaystyle \text{Let } x-\frac{1}{x}=t \Rightarrow \left(1+\frac{1}{x^{2}}\right)dx=dt
\displaystyle \therefore I_{1}=\frac{1}{2}\int \frac{dt}{t^{2}+2}
\displaystyle =\frac{1}{2}\cdot \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t}{\sqrt{2}}\right)+C_{1}
\displaystyle =\frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)+C_{1}
\displaystyle I_{2}=\frac{1}{2}\int \frac{1-\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}}\,dx
\displaystyle \text{Let } x+\frac{1}{x}=t \Rightarrow \left(1-\frac{1}{x^{2}}\right)dx=dt
\displaystyle \therefore I_{2}=\frac{1}{2}\int \frac{dt}{t^{2}-2}
\displaystyle =\frac{1}{4\sqrt{2}}\log\left|\frac{t-\sqrt{2}}{t+\sqrt{2}}\right|+C_{2}
\displaystyle =\frac{1}{4\sqrt{2}}\log\left|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right|+C_{2}
\displaystyle \therefore I=I_{1}+I_{2}
\displaystyle =\frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)+\frac{1}{4\sqrt{2}}\log\left|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right|+C
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\displaystyle \textbf{Question 22: } \text{Evaluate } \int \frac{1}{\sin^{4}x+\cos^{4}x}\,dx.   
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{1}{\sin^{4}x+\cos^{4}x}\,dx=\int \frac{\sec^{4}x}{1+\tan^{4}x}\,dx
\displaystyle \text{On putting } \tan x=t \Rightarrow \sec^{2}x\,dx=dt
\displaystyle \therefore I=\int \frac{1+t^{2}}{t^{4}+1}\,dt
\displaystyle =\int \frac{1+\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}}\,dt
\displaystyle =\int \frac{1+\frac{1}{t^{2}}}{\left(t-\frac{1}{t}\right)^{2}+2}\,dt
\displaystyle \text{Let } t-\frac{1}{t}=v \Rightarrow \left(1+\frac{1}{t^{2}}\right)dt=dv
\displaystyle \therefore I=\int \frac{dv}{v^{2}+2}=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{v}{\sqrt{2}}\right)+C
\displaystyle =\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)+C
\displaystyle =\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\tan x-\cot x}{\sqrt{2}}\right)+C
\\

\displaystyle \textbf{Question 23: } \text{Evaluate } \int \frac{dx}{\sin x+\sin 2x}. \quad   
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{dx}{\sin x+\sin 2x}=\int \frac{dx}{\sin x+2\sin x\cos x}
\displaystyle =\int \frac{dx}{\sin x(1+2\cos x)}
\displaystyle \text{On putting } \cos x=t \Rightarrow -\sin x\,dx=dt
\displaystyle \therefore I=-\int \frac{dt}{(1-t^{2})(1+2t)}
\displaystyle =-\int \frac{dt}{(1+t)(1-t)(1+2t)}
\displaystyle \text{Let } \frac{-1}{(1+t)(1-t)(1+2t)}=\frac{A}{1+t}+\frac{B}{1-t}+\frac{C}{1+2t}
\displaystyle \Rightarrow -1=A(1-t)(1+2t)+B(1+t)(1+2t)+C(1+t)(1-t)
\displaystyle \text{On putting } t=1,\ B=-\frac{1}{6};\ t=-1,\ A=\frac{1}{2};\ t=-\frac{1}{2},\ C=-\frac{4}{3}
\displaystyle \therefore I=\int \left[\frac{1}{2(1+t)}-\frac{1}{6(1-t)}-\frac{4}{3(1+2t)}\right]dt
\displaystyle =\frac{1}{2}\log|1+t|+\frac{1}{6}\log|1-t|-\frac{2}{3}\log|1+2t|+C
\displaystyle \Rightarrow I=\frac{1}{2}\log|1+\cos x|+\frac{1}{6}\log|1-\cos x|-\frac{2}{3}\log|1+2\cos x|+C
\\

\displaystyle \textbf{Question 24: } \text{Evaluate } \int \tan^{-1}\sqrt{x}\,dx. 
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \tan^{-1}\sqrt{x}\,dx
\displaystyle \text{On putting } x=\tan^{2}\theta \Rightarrow dx=2\tan\theta\sec^{2}\theta\,d\theta
\displaystyle \therefore I=2\int \theta\tan\theta\sec^{2}\theta\,d\theta
\displaystyle =2\int \theta(\sec^{2}\theta-1)\,d\theta
\displaystyle =2\left[\theta\tan\theta-\int \tan\theta\,d\theta\right]
\displaystyle =2[\theta\tan\theta+\log|\cos\theta|]+C
\displaystyle =2\left[\tan^{-1}\sqrt{x}\cdot \sqrt{x}+\log\left|\frac{1}{\sqrt{1+x}}\right|\right]+C
\displaystyle =(x+1)\tan^{-1}\sqrt{x}-\sqrt{x}+C
\\

\displaystyle \textbf{Question 25: } \text{Evaluate } \int \frac{3x+5}{x^{3}-x^{2}-x+1}\,dx.   
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{3x+5}{x^{3}-x^{2}-x+1}\,dx
\displaystyle \Rightarrow I=\int \frac{3x+5}{(x+1)(x-1)^{2}}\,dx
\displaystyle \text{Here, integrand is a proper rational function, so using partial fraction, we write as}
\displaystyle \frac{3x+5}{(x+1)(x-1)^{2}}=\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{(x-1)^{2}}
\displaystyle \Rightarrow 3x+5=A(x-1)^{2}+B(x+1)(x-1)+C(x+1) \quad (i)
\displaystyle \text{On putting } x=1 \text{ in Eq. (i), we get } 8=2C \Rightarrow C=4
\displaystyle \text{On putting } x=-1 \text{ in Eq. (i), we get } 2=4A \Rightarrow A=\frac{1}{2}
\displaystyle \text{On putting } x=0 \text{ in Eq. (i), we get } 5=A-B+C
\displaystyle \Rightarrow 5=\frac{1}{2}-B+4 \Rightarrow B=-\frac{1}{2}
\displaystyle \therefore I=\frac{1}{2}\int \frac{dx}{x+1}-\frac{1}{2}\int \frac{dx}{x-1}+4\int \frac{dx}{(x-1)^{2}}
\displaystyle =\frac{1}{2}\log|x+1|-\frac{1}{2}\log|x-1|-\frac{4}{x-1}+C
\displaystyle \Rightarrow I=\frac{1}{2}\log\left|\frac{x+1}{x-1}\right|-\frac{4}{x-1}+C
\\

\displaystyle \textbf{Question 26: } \text{Evaluate } \int \frac{9x^{3}+63x^{2}}{x^{4}+14x^{3}-9x^{2}}\,dx. \quad  
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{9x^{3}+63x^{2}}{x^{4}+14x^{3}-9x^{2}}\,dx
\displaystyle =\int \frac{9x^{2}(x+7)}{x^{2}(x^{2}+14x-9)}\,dx=\int \frac{9(x+7)}{x^{2}+14x-9}\,dx
\displaystyle \text{Let } x^{2}+14x-9=t \Rightarrow (2x+14)dx=dt
\displaystyle \Rightarrow (x+7)dx=\frac{dt}{2}
\displaystyle \therefore I=\frac{9}{2}\int \frac{dt}{t}=\frac{9}{2}\log|t|+C
\displaystyle =\frac{9}{2}\log|x^{2}+14x-9|+C
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