\displaystyle \text{Evaluate the following integrals:}

\displaystyle \textbf{Question 1: }~\int \frac{x^2+1}{x^4+x^2+1}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int\frac{x^2+1}{x^4+x^2+1}\,dx
\displaystyle \text{Dividing numerator and denominator by }x^2
\displaystyle I=\int\frac{1+\frac{1}{x^2}}{x^2+1+\frac{1}{x^2}}\,dx
\displaystyle I=\int\frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2+3}\,dx
\displaystyle \text{Putting }x-\frac{1}{x}=t
\displaystyle \left(1+\frac{1}{x^2}\right)dx=dt
\displaystyle I=\int\frac{dt}{t^2+3}
\displaystyle I=\int\frac{dt}{t^2+(\sqrt{3})^2}
\displaystyle I=\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{t}{\sqrt{3}}\right)+C
\displaystyle I=\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right)+C
\displaystyle I=\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{x^2-1}{\sqrt{3}x}\right)+C

\displaystyle \textbf{Question 2: }~\int \sqrt{\cot\theta}\,d\theta.
\displaystyle \text{Answer:}
\displaystyle I=\int\sqrt{\cot\theta}\,d\theta
\displaystyle \text{Putting }\cot\theta=t^2
\displaystyle -\mathrm{cosec}^2\theta\,d\theta=2t\,dt
\displaystyle d\theta=-\frac{2t\,dt}{1+\cot^2\theta}
\displaystyle d\theta=-\frac{2t\,dt}{1+t^4}
\displaystyle I=\int t\left(-\frac{2t\,dt}{1+t^4}\right)
\displaystyle I=-\int\frac{2t^2}{1+t^4}\,dt
\displaystyle I=-\int\frac{t^2+1+t^2-1}{t^4+1}\,dt
\displaystyle I=-\int\frac{t^2+1}{t^4+1}\,dt-\int\frac{t^2-1}{t^4+1}\,dt
\displaystyle \text{Dividing numerator and denominator by }t^2
\displaystyle I=-\int\frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}\,dt-\int\frac{1-\frac{1}{t^2}}{t^2+\frac{1}{t^2}}\,dt
\displaystyle I=-\int\frac{1+\frac{1}{t^2}}{\left(t-\frac{1}{t}\right)^2+2}\,dt-\int\frac{1-\frac{1}{t^2}}{\left(t+\frac{1}{t}\right)^2-2}\,dt
\displaystyle \text{Putting }t-\frac{1}{t}=p
\displaystyle \left(1+\frac{1}{t^2}\right)dt=dp
\displaystyle \text{Putting }t+\frac{1}{t}=q
\displaystyle \left(1-\frac{1}{t^2}\right)dt=dq
\displaystyle I=-\int\frac{dp}{p^2+(\sqrt{2})^2}-\int\frac{dq}{q^2-(\sqrt{2})^2}
\displaystyle I=-\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{p}{\sqrt{2}}\right)-\frac{1}{2\sqrt{2}}\log\left|\frac{q-\sqrt{2}}{q+\sqrt{2}}\right|+C
\displaystyle I=-\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{2}}\right)-\frac{1}{2\sqrt{2}}\log\left|\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\right|+C
\displaystyle I=-\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t^2-1}{\sqrt{2}t}\right)-\frac{1}{2\sqrt{2}}\log\left|\frac{t^2+1-\sqrt{2}t}{t^2+1+\sqrt{2}t}\right|+C
\displaystyle I=-\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\cot\theta-1}{2\sqrt{\cot\theta}}\right)-\frac{1}{2\sqrt{2}}\log\left|\frac{\cot\theta+1-\sqrt{2\cot\theta}}{\cot\theta+1+\sqrt{2\cot\theta}}\right|+C

\displaystyle \textbf{Question 3: }~\int \frac{x^2+9}{x^4+81}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int\frac{x^2+9}{x^4+81}\,dx
\displaystyle \text{Dividing numerator and denominator by }x^2
\displaystyle I=\int\frac{1+\frac{9}{x^2}}{x^2+\frac{81}{x^2}}\,dx
\displaystyle I=\int\frac{1+\frac{9}{x^2}}{\left(x-\frac{9}{x}\right)^2+18}\,dx
\displaystyle \text{Putting }x-\frac{9}{x}=t
\displaystyle \left(1+\frac{9}{x^2}\right)dx=dt
\displaystyle I=\int\frac{dt}{t^2+18}
\displaystyle I=\int\frac{dt}{t^2+(3\sqrt{2})^2}
\displaystyle I=\frac{1}{3\sqrt{2}}\tan^{-1}\left(\frac{t}{3\sqrt{2}}\right)+C
\displaystyle I=\frac{1}{3\sqrt{2}}\tan^{-1}\left(\frac{x-\frac{9}{x}}{3\sqrt{2}}\right)+C
\displaystyle I=\frac{1}{3\sqrt{2}}\tan^{-1}\left(\frac{x^2-9}{3\sqrt{2}\,x}\right)+C

\displaystyle \textbf{Question 4: }~\int \frac{1}{x^4+x^2+1}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int\frac{dx}{x^4+x^2+1}
\displaystyle I=\frac12\int\frac{2\,dx}{x^4+x^2+1}
\displaystyle I=\frac12\int\frac{(x^2+1)-(x^2-1)}{x^4+x^2+1}\,dx
\displaystyle I=\frac12\int\frac{x^2+1}{x^4+x^2+1}\,dx-\frac12\int\frac{x^2-1}{x^4+x^2+1}\,dx
\displaystyle \text{Dividing numerator and denominator by }x^2
\displaystyle I=\frac12\int\frac{1+\frac1{x^2}}{x^2+1+\frac1{x^2}}\,dx-\frac12\int\frac{1-\frac1{x^2}}{x^2+1+\frac1{x^2}}\,dx
\displaystyle I=\frac12\int\frac{1+\frac1{x^2}}{\left(x-\frac1x\right)^2+3}\,dx-\frac12\int\frac{1-\frac1{x^2}}{\left(x+\frac1x\right)^2-1}\,dx
\displaystyle \text{Putting }x-\frac1x=t
\displaystyle \left(1+\frac1{x^2}\right)dx=dt
\displaystyle \text{Putting }x+\frac1x=p
\displaystyle \left(1-\frac1{x^2}\right)dx=dp
\displaystyle I=\frac12\int\frac{dt}{t^2+(\sqrt3)^2}-\frac12\int\frac{dp}{p^2-1^2}
\displaystyle I=\frac{1}{2\sqrt3}\tan^{-1}\left(\frac{t}{\sqrt3}\right)-\frac14\log\left|\frac{p-1}{p+1}\right|+C
\displaystyle I=\frac{1}{2\sqrt3}\tan^{-1}\left(\frac{x-\frac1x}{\sqrt3}\right)-\frac14\log\left|\frac{x+\frac1x-1}{x+\frac1x+1}\right|+C
\displaystyle I=\frac{1}{2\sqrt3}\tan^{-1}\left(\frac{x^2-1}{x\sqrt3}\right)-\frac14\log\left|\frac{x^2-x+1}{x^2+x+1}\right|+C

\displaystyle \textbf{Question 5: }~\int \frac{x^2-3x+1}{x^4+x^2+1}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int\frac{x^2-3x+1}{x^4+x^2+1}\,dx
\displaystyle I=\int\frac{x^2+1}{x^4+x^2+1}\,dx-3\int\frac{x\,dx}{x^4+x^2+1}
\displaystyle \text{Let }I_1=\int\frac{x^2+1}{x^4+x^2+1}\,dx,\;I_2=\int\frac{x\,dx}{x^4+x^2+1}
\displaystyle I_1=\int\frac{x^2+1}{x^4+x^2+1}\,dx
\displaystyle \text{Dividing numerator and denominator by }x^2
\displaystyle I_1=\int\frac{1+\frac1{x^2}}{x^2+1+\frac1{x^2}}\,dx
\displaystyle I_1=\int\frac{1+\frac1{x^2}}{\left(x-\frac1x\right)^2+3}\,dx
\displaystyle \text{Putting }x-\frac1x=t
\displaystyle \left(1+\frac1{x^2}\right)dx=dt
\displaystyle I_1=\int\frac{dt}{t^2+(\sqrt3)^2}
\displaystyle I_1=\frac{1}{\sqrt3}\tan^{-1}\left(\frac{t}{\sqrt3}\right)+C_1
\displaystyle I_1=\frac{1}{\sqrt3}\tan^{-1}\left(\frac{x-\frac1x}{\sqrt3}\right)+C_1
\displaystyle I_2=\int\frac{x\,dx}{x^4+x^2+1}
\displaystyle \text{Putting }x^2=t
\displaystyle 2x\,dx=dt
\displaystyle x\,dx=\frac{dt}{2}
\displaystyle I_2=\frac12\int\frac{dt}{t^2+t+1}
\displaystyle I_2=\frac12\int\frac{dt}{\left(t+\frac12\right)^2+\left(\frac{\sqrt3}{2}\right)^2}
\displaystyle I_2=\frac{1}{\sqrt3}\tan^{-1}\left(\frac{2t+1}{\sqrt3}\right)+C_2
\displaystyle I_2=\frac{1}{\sqrt3}\tan^{-1}\left(\frac{2x^2+1}{\sqrt3}\right)+C_2
\displaystyle I=\frac{1}{\sqrt3}\tan^{-1}\left(\frac{x^2-1}{x\sqrt3}\right)-\sqrt3\tan^{-1}\left(\frac{2x^2+1}{\sqrt3}\right)+C

\displaystyle \textbf{Question 6: }~\int \frac{x^2+1}{x^4-x^2+1}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int\frac{x^2+1}{x^4-x^2+1}\,dx
\displaystyle \text{Dividing numerator and denominator by }x^2
\displaystyle I=\int\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}-1}\,dx
\displaystyle I=\int\frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2+1}\,dx
\displaystyle \text{Putting }x-\frac{1}{x}=t
\displaystyle \left(1+\frac{1}{x^2}\right)dx=dt
\displaystyle I=\int\frac{dt}{t^2+1}
\displaystyle I=\tan^{-1}t+C
\displaystyle I=\tan^{-1}\left(x-\frac{1}{x}\right)+C
\displaystyle I=\tan^{-1}\left(\frac{x^2-1}{x}\right)+C

\displaystyle \textbf{Question 7: }~\int \frac{x^2-1}{x^4+1}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int\frac{x^2-1}{x^4+1}\,dx
\displaystyle \text{Dividing numerator and denominator by }x^2
\displaystyle I=\int\frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}}\,dx
\displaystyle I=\int\frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2-2}\,dx
\displaystyle \text{Putting }x+\frac{1}{x}=t
\displaystyle \left(1-\frac{1}{x^2}\right)dx=dt
\displaystyle I=\int\frac{dt}{t^2-(\sqrt{2})^2}
\displaystyle I=\frac{1}{2\sqrt{2}}\log\left|\frac{t-\sqrt{2}}{t+\sqrt{2}}\right|+C
\displaystyle I=\frac{1}{2\sqrt{2}}\log\left|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right|+C
\displaystyle I=\frac{1}{2\sqrt{2}}\log\left|\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right|+C

\displaystyle \textbf{Question 8: }~\int \frac{x^2+1}{x^4+7x^2+1}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int\frac{x^2+1}{x^4+7x^2+1}\,dx
\displaystyle \text{Dividing numerator and denominator by }x^2
\displaystyle I=\int\frac{1+\frac{1}{x^2}}{x^2+7+\frac{1}{x^2}}\,dx
\displaystyle I=\int\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}-2+9}\,dx
\displaystyle I=\int\frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2+3^2}\,dx
\displaystyle \text{Putting }x-\frac{1}{x}=t
\displaystyle \left(1+\frac{1}{x^2}\right)dx=dt
\displaystyle I=\int\frac{dt}{t^2+3^2}
\displaystyle I=\frac{1}{3}\tan^{-1}\left(\frac{t}{3}\right)+C
\displaystyle I=\frac{1}{3}\tan^{-1}\left(\frac{x-\frac{1}{x}}{3}\right)+C
\displaystyle I=\frac{1}{3}\tan^{-1}\left(\frac{x^2-1}{3x}\right)+C

\displaystyle \textbf{Question 9: }~\int \frac{(x-1)^2}{x^4+x^2+1}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int\frac{(x-1)^2}{x^4+x^2+1}\,dx
\displaystyle I=\int\frac{x^2-2x+1}{x^4+x^2+1}\,dx
\displaystyle I=\int\frac{x^2+1}{x^4+x^2+1}\,dx-\int\frac{2x}{x^4+x^2+1}\,dx
\displaystyle I=I_1-I_2
\displaystyle \text{where }I_1=\int\frac{x^2+1}{x^4+x^2+1}\,dx,\;I_2=\int\frac{2x}{x^4+x^2+1}\,dx
\displaystyle I_1=\int\frac{x^2+1}{x^4+x^2+1}\,dx
\displaystyle \text{Dividing numerator and denominator by }x^2
\displaystyle I_1=\int\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}+1}\,dx
\displaystyle I_1=\int\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}-2+3}\,dx
\displaystyle I_1=\int\frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2+3}\,dx
\displaystyle \text{Putting }x-\frac{1}{x}=t
\displaystyle \left(1+\frac{1}{x^2}\right)dx=dt
\displaystyle I_1=\int\frac{dt}{t^2+(\sqrt3)^2}
\displaystyle I_1=\frac{1}{\sqrt3}\tan^{-1}\left(\frac{t}{\sqrt3}\right)+C_1
\displaystyle I_1=\frac{1}{\sqrt3}\tan^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt3}\right)+C_1
\displaystyle I_2=\int\frac{2x}{x^4+x^2+1}\,dx
\displaystyle \text{Putting }x^2=t
\displaystyle 2x\,dx=dt
\displaystyle I_2=\int\frac{dt}{t^2+t+1}
\displaystyle I_2=\int\frac{dt}{\left(t+\frac12\right)^2+\left(\frac{\sqrt3}{2}\right)^2}
\displaystyle I_2=\frac{2}{\sqrt3}\tan^{-1}\left(\frac{t+\frac12}{\frac{\sqrt3}{2}}\right)+C_2
\displaystyle I_2=\frac{2}{\sqrt3}\tan^{-1}\left(\frac{2t+1}{\sqrt3}\right)+C_2
\displaystyle I_2=\frac{2}{\sqrt3}\tan^{-1}\left(\frac{2x^2+1}{\sqrt3}\right)+C_2
\displaystyle I=\frac{1}{\sqrt3}\tan^{-1}\left(\frac{x^2-1}{x\sqrt3}\right)-\frac{2}{\sqrt3}\tan^{-1}\left(\frac{2x^2+1}{\sqrt3}\right)+C

\displaystyle \textbf{Question 10: }~\int \frac{1}{x^4+3x^2+1}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int\frac{dx}{x^4+3x^2+1}
\displaystyle I=\frac12\int\frac{2\,dx}{x^4+3x^2+1}
\displaystyle I=\frac12\int\frac{(x^2+1)-(x^2-1)}{x^4+3x^2+1}\,dx
\displaystyle I=\frac12\int\frac{x^2+1}{x^4+3x^2+1}\,dx-\frac12\int\frac{x^2-1}{x^4+3x^2+1}\,dx
\displaystyle \text{Dividing numerator and denominator by }x^2
\displaystyle I=\frac12\int\frac{1+\frac1{x^2}}{x^2+\frac1{x^2}+3}\,dx-\frac12\int\frac{1-\frac1{x^2}}{x^2+\frac1{x^2}+3}\,dx
\displaystyle I=\frac12\int\frac{1+\frac1{x^2}}{\left(x-\frac1x\right)^2+5}\,dx-\frac12\int\frac{1-\frac1{x^2}}{\left(x+\frac1x\right)^2+1}\,dx
\displaystyle \text{Putting }x-\frac1x=t
\displaystyle \left(1+\frac1{x^2}\right)dx=dt
\displaystyle \text{Putting }x+\frac1x=p
\displaystyle \left(1-\frac1{x^2}\right)dx=dp
\displaystyle I=\frac12\int\frac{dt}{t^2+(\sqrt5)^2}-\frac12\int\frac{dp}{p^2+1^2}
\displaystyle I=\frac{1}{2\sqrt5}\tan^{-1}\left(\frac{t}{\sqrt5}\right)-\frac12\tan^{-1}(p)+C
\displaystyle I=\frac{1}{2\sqrt5}\tan^{-1}\left(\frac{x-\frac1x}{\sqrt5}\right)-\frac12\tan^{-1}\left(x+\frac1x\right)+C
\displaystyle I=\frac{1}{2\sqrt5}\tan^{-1}\left(\frac{x^2-1}{\sqrt5\,x}\right)-\frac12\tan^{-1}\left(\frac{x^2+1}{x}\right)+C

\displaystyle \textbf{Question 11: }~\int \frac{1}{\sin^4 x+\sin^2 x\cos^2 x+\cos^4 x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{dx}{\sin^4x+\sin^2x\cos^2x+\cos^4x}
\displaystyle I=\int\frac{dx}{(\sin^2x+\cos^2x)^2-\sin^2x\cos^2x}
\displaystyle I=\int\frac{dx}{1-\sin^2x\cos^2x}
\displaystyle I=\int\frac{dx}{1-\frac{\tan^2x}{\sec^4x}}
\displaystyle I=\int\frac{\sec^4x}{\sec^4x-\tan^2x}\,dx
\displaystyle I=\int\frac{\sec^2x(1+\tan^2x)}{\sec^4x-\tan^2x}\,dx
\displaystyle \text{Let }\tan x=t
\displaystyle \sec^2x\,dx=dt
\displaystyle I=\int\frac{1+t^2}{(1+t^2)^2-t^2}\,dt
\displaystyle I=\int\frac{1+t^2}{t^4+t^2+1}\,dt
\displaystyle I=\int\frac{1+\frac{1}{t^2}}{t^2+1+\frac{1}{t^2}}\,dt
\displaystyle I=\int\frac{1+\frac{1}{t^2}}{(t-\frac{1}{t})^2+3}\,dt
\displaystyle \text{Let }t-\frac{1}{t}=u
\displaystyle \left(1+\frac{1}{t^2}\right)dt=du
\displaystyle I=\int\frac{du}{u^2+3}
\displaystyle I=\frac{1}{\sqrt3}\tan^{-1}\left(\frac{u}{\sqrt3}\right)+C
\displaystyle I=\frac{1}{\sqrt3}\tan^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt3}\right)+C
\displaystyle I=\frac{1}{\sqrt3}\tan^{-1}\left(\frac{\tan x-\cot x}{\sqrt3}\right)+C

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