\displaystyle \textbf{Question 1: }~\int \frac{\sec^2 x}{1-\tan^2 x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let }t=\tan x\Rightarrow dt=\sec^2 x\,dx.
\displaystyle \int \frac{\sec^2 x}{1-\tan^2 x}\,dx=\int \frac{dt}{1-t^2}=\frac12\log\left|\frac{1+t}{1-t}\right|+C.
\displaystyle =\frac12\log\left|\frac{1+\tan x}{1-\tan x}\right|+C.
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\displaystyle \textbf{Question 2: }~\int \frac{e^x}{1+e^{2x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let }t=e^x\Rightarrow dt=e^x\,dx.
\displaystyle \int \frac{e^x}{1+e^{2x}}\,dx=\int \frac{dt}{1+t^2}=\tan^{-1}(t)+C=\tan^{-1}(e^x)+C.
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\displaystyle \textbf{Question 3: }~\int \frac{\cos x}{\sin^2 x+4\sin x+5}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let }t=\sin x\Rightarrow dt=\cos x\,dx.
\displaystyle \int \frac{\cos x}{\sin^2 x+4\sin x+5}\,dx=\int \frac{dt}{t^2+4t+5}=\int \frac{dt}{(t+2)^2+1}.
\displaystyle =\tan^{-1}(t+2)+C=\tan^{-1}(\sin x+2)+C.
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\displaystyle \textbf{Question 4: }~\int \frac{e^x}{e^{2x}+5e^x+6}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let }t=e^x\Rightarrow dt=e^x\,dx.
\displaystyle \int \frac{e^x}{e^{2x}+5e^x+6}\,dx=\int \frac{dt}{t^2+5t+6}=\int \frac{dt}{(t+2)(t+3)}.
\displaystyle =\int\left(\frac{1}{t+2}-\frac{1}{t+3}\right)dt=\log|t+2|-\log|t+3|+C.
\displaystyle =\log\left|\frac{e^x+2}{e^x+3}\right|+C.
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\displaystyle \textbf{Question 5: }~\int \frac{e^{3x}}{4e^{6x}-9}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let }t=e^{3x}\Rightarrow dt=3e^{3x}\,dx\Rightarrow e^{3x}\,dx=\frac{dt}{3}.
\displaystyle \int \frac{e^{3x}}{4e^{6x}-9}\,dx=\frac13\int \frac{dt}{4t^2-9}.
\displaystyle =\frac13\cdot\frac{1}{12}\log\left|\frac{2t-3}{2t+3}\right|+C=\frac{1}{36}\log\left|\frac{2t-3}{2t+3}\right|+C.
\displaystyle =\frac{1}{36}\log\left|\frac{2e^{3x}-3}{2e^{3x}+3}\right|+C.
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\displaystyle \textbf{Question 6: }~\int \frac{1}{e^x+e^{-x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \int \frac{1}{e^x+e^{-x}}\,dx=\int \frac{e^x}{e^{2x}+1}\,dx.
\displaystyle \text{Let }t=e^x\Rightarrow dt=e^x\,dx.
\displaystyle \int \frac{e^x}{e^{2x}+1}\,dx=\int \frac{dt}{t^2+1}=\tan^{-1}(t)+C=\tan^{-1}(e^x)+C.
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\displaystyle \textbf{Question 7: }~\int \frac{x}{x^4+2x^2+3}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let }t=x^2\Rightarrow dt=2x\,dx\Rightarrow x\,dx=\frac{dt}{2}.
\displaystyle \int \frac{x}{x^4+2x^2+3}\,dx=\frac12\int \frac{dt}{t^2+2t+3}=\frac12\int \frac{dt}{(t+1)^2+2}.
\displaystyle =\frac{1}{2\sqrt2}\tan^{-1}\left(\frac{t+1}{\sqrt2}\right)+C=\frac{1}{2\sqrt2}\tan^{-1}\left(\frac{x^2+1}{\sqrt2}\right)+C.
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\displaystyle \textbf{Question 8: }~\int \frac{3x^5}{1+x^{12}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let }t=x^6\Rightarrow dt=6x^5\,dx\Rightarrow x^5\,dx=\frac{dt}{6}.
\displaystyle \int \frac{3x^5}{1+x^{12}}\,dx=\int \frac{3x^5}{1+(x^6)^2}\,dx=\frac{3}{6}\int \frac{dt}{1+t^2}.
\displaystyle =\frac12\tan^{-1}(t)+C=\frac12\tan^{-1}(x^6)+C.
\\
\displaystyle \textbf{Question 9: }~\int \frac{x^2}{x^6-a^6}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let }t=x^3\Rightarrow dt=3x^2\,dx\Rightarrow x^2\,dx=\frac{dt}{3}.
\displaystyle \int \frac{x^2}{x^6-a^6}\,dx=\frac13\int \frac{dt}{t^2-a^6}=\frac13\int \frac{dt}{(t-a^3)(t+a^3)}.
\displaystyle =\frac{1}{6a^3}\log\left|\frac{t-a^3}{t+a^3}\right|+C=\frac{1}{6a^3}\log\left|\frac{x^3-a^3}{x^3+a^3}\right|+C.
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\displaystyle \textbf{Question 10: }~\int \frac{x^2}{x^6+a^6}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let }t=x^3\Rightarrow dt=3x^2\,dx\Rightarrow x^2\,dx=\frac{dt}{3}.
\displaystyle \int \frac{x^2}{x^6+a^6}\,dx=\frac13\int \frac{dt}{t^2+a^6}.
\displaystyle =\frac13\cdot\frac{1}{a^3}\tan^{-1}\left(\frac{t}{a^3}\right)+C=\frac{1}{3a^3}\tan^{-1}\left(\frac{x^3}{a^3}\right)+C.
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\displaystyle \textbf{Question 11: }~\int \frac{1}{x(x^6+1)}\,dx.
\displaystyle \text{Answer:}
\displaystyle \int \frac{1}{x(x^6+1)}\,dx=\int \frac{1}{x}\cdot\frac{1}{1+x^6}\,dx.
\displaystyle \text{Let }t=x^6\Rightarrow \frac{dt}{6t}=\frac{dx}{x}.
\displaystyle =\frac16\int \frac{1}{t(1+t)}\,dt=\frac16\int\left(\frac{1}{t}-\frac{1}{1+t}\right)\!dt.
\displaystyle =\frac16\log|t|-\frac16\log|1+t|+C=\frac16\log\left|\frac{x^6}{1+x^6}\right|+C.
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\displaystyle \textbf{Question 12: }~\int \frac{x}{x^4-x^2+1}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let }t=x^2\Rightarrow dt=2x\,dx\Rightarrow x\,dx=\frac{dt}{2}.
\displaystyle \int \frac{x}{x^4-x^2+1}\,dx=\frac12\int \frac{dt}{t^2-t+1}=\frac12\int \frac{dt}{\left(t-\frac12\right)^2+\left(\frac{\sqrt3}{2}\right)^2}.
\displaystyle =\frac{1}{\sqrt3}\tan^{-1}\left(\frac{2t-1}{\sqrt3}\right)+C =\frac{1}{\sqrt3}\tan^{-1}\left(\frac{2x^2-1}{\sqrt3}\right)+C.
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\displaystyle \textbf{Question 13: }~\int \frac{x}{3x^4-18x^2+11}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let }t=x^2\Rightarrow dt=2x\,dx\Rightarrow x\,dx=\frac{dt}{2}.
\displaystyle \int \frac{x}{3x^4-18x^2+11}\,dx=\frac12\int \frac{dt}{3t^2-18t+11}.
\displaystyle 3t^2-18t+11=3(t-3)^2-16.
\displaystyle \therefore \frac12\int \frac{dt}{3(t-3)^2-16}=\frac16\int \frac{du}{u^2-\left(\frac{4}{\sqrt3}\right)^2}\quad(u=t-3).
\displaystyle =\frac{1}{16\sqrt3}\log\left|\frac{u-\frac{4}{\sqrt3}}{u+\frac{4}{\sqrt3}}\right|+C =\frac{1}{16\sqrt3}\log\left|\frac{\sqrt3u-4}{\sqrt3u+4}\right|+C.
\displaystyle =\frac{1}{16\sqrt3}\log\left|\frac{\sqrt3(x^2-3)-4}{\sqrt3(x^2-3)+4}\right|+C.
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\displaystyle \textbf{Question 14: }~\int \frac{e^x}{(1+e^x)(2+e^x)}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let }t=e^x\Rightarrow dt=e^x\,dx.
\displaystyle \int \frac{e^x}{(1+e^x)(2+e^x)}\,dx=\int \frac{dt}{(1+t)(2+t)}.
\displaystyle =\int\left(\frac{1}{1+t}-\frac{1}{2+t}\right)\!dt=\log|1+t|-\log|2+t|+C.
\displaystyle =\log\left|\frac{1+e^x}{2+e^x}\right|+C.
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\displaystyle \textbf{Question 15: }~\int \frac{1}{\cos x+\mathrm{cosec}\,x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{1}{\cos x+\mathrm{cosec}\,x}\,dx
\displaystyle =\int \frac{\sin x}{1+\cos x+\sin x}\,dx
\displaystyle =\int \frac{2\sin x}{2+2\cos x+\sin x}\,dx
\displaystyle =\int \frac{\sin x+\cos x+\sin x-\cos x}{2+2\cos x+\sin x}\,dx
\displaystyle =\int \frac{\sin x+\cos x}{2+2\cos x+\sin x}\,dx+\int \frac{\sin x-\cos x}{2+2\cos x+\sin x}\,dx
\displaystyle =\int \frac{\sin x+\cos x}{3-\sin^{2}x-\cos^{2}x+2\cos x\sin x}\,dx+\int \frac{\sin x-\cos x}{1+\sin^{2}x+\cos^{2}x+2\cos x\sin x}\,dx
\displaystyle =\int \frac{\sin x+\cos x}{3-(\sin x-\cos x)^{2}}\,dx+\int \frac{\sin x-\cos x}{1+(\sin x+\cos x)^{2}}\,dx
\displaystyle \text{where, } I_{1}=\int \frac{\sin x+\cos x}{3-(\sin x-\cos x)^{2}}\,dx \text{ and } I_{2}=\int \frac{\sin x-\cos x}{1+(\sin x+\cos x)^{2}}\,dx
\displaystyle \text{Now,}
\displaystyle I_{1}=\int \frac{\sin x+\cos x}{3-(\sin x-\cos x)^{2}}\,dx
\displaystyle \text{Let } (\sin x-\cos x)=t
\displaystyle \text{On differentiating both sides, we get}
\displaystyle (\cos x+\sin x)\,dx=dt
\displaystyle \therefore I_{1}=\int \frac{1}{3-t^{2}}\,dt
\displaystyle =\frac{1}{2\sqrt{3}}\log\left|\frac{\sqrt{3}+t}{\sqrt{3}-t}\right|+c_{1}
\displaystyle =\frac{1}{2\sqrt{3}}\log\left|\frac{\sqrt{3}+\sin x-\cos x}{\sqrt{3}-(\sin x-\cos x)}\right|+c_{1}\ldots(2)
\displaystyle \text{Now,}
\displaystyle I_{2}=\int \frac{\sin x-\cos x}{1+(\sin x+\cos x)^{2}}\,dx
\displaystyle \text{Let } (\sin x+\cos x)=t
\displaystyle \text{On differentiating both sides, we get}
\displaystyle (\cos x-\sin x)\,dx=dt
\displaystyle \therefore I_{2}=-\int \frac{1}{1+t^{2}}\,dt
\displaystyle =-\tan^{-1}t+c_{2}
\displaystyle =-\tan^{-1}(\sin x+\cos x)+c_{2}\ldots(3)
\displaystyle \text{On substituting (2) and (3) in (1), we get}
\displaystyle I=\frac{1}{2\sqrt{3}}\log\left|\frac{\sqrt{3}+\sin x-\cos x}{\sqrt{3}-\sin x+\cos x}\right|-\tan^{-1}(\sin x+\cos x)+c
\displaystyle \text{Hence, } \int \frac{1}{\cos x+\mathrm{cosec}\,x}\,dx=\frac{1}{2\sqrt{3}}\log\left|\frac{\sqrt{3}+\sin x-\cos x}{\sqrt{3}-\sin x+\cos x}\right|-\tan^{-1}(\sin x+\cos x)+c

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