\({tan^{11}\ x\over 11}+{tan^9\ x\over 9}+c\)

\({2\over 3\sqrt2}tan^{-1}{x\over\sqrt2}+{1\over 6}log|{x-1\over x+1}|+c\)

\(-{1\over 3}({1+{1\over x^2}})^{3/2}+c\)

\(l={\pi\over3}\)

\(= tan^{-1}e-{\pi\over4}\)

\(I=\int \sqrt{\frac{1+x}{1-x}} d x=\int \sqrt{\frac{(1+x)(1+x)}{(1-x)(1+x)}} d x\) 
\(=\int \frac{1+x}{\sqrt{1-x^{2}}} d x \)
\(=\int \frac{d x}{\sqrt{1-x^{2}}}+\int \frac{x}{\sqrt{1-x^{2}}} d x \)
Put \(1-x^{2}=t\) in second integral
[Ans.\(sin ^{-1}(x)-\sqrt{1-x^{2}}+C\)]

\(\frac{2 x-1}{(x-1)(x+2)(x-3)}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x-3}\) 
\(\log \left|\frac{\sqrt{x-3}}{(x-1)^{1 / 6}(x+2)^{1 / 3}}\right|+C\)

\( \frac{(x-1)}{2} \sqrt{5-2 x+x^{2}}+2 \log \mid(x-1) +\sqrt{5-2 x+x^{2}} \mid+C \)

\(\frac{-9}{2}\)

Let \(I=\int_{0}^{1} x\left(\tan ^{-1} x\right)^{2} d x\) 
Now, put \(x=\tan \theta \Rightarrow d x=\sec ^{2} \theta d \theta\) 
Upper limit When \(x=1, \tan \theta=1 \Rightarrow \theta=\tan ^{-1}(1)=\frac{\pi}{4}\) 
Lower limit When \(x=0, \tan \theta=0 \Rightarrow \theta=\tan ^{-1}(0)=0\) 
Now, from Eq. (i), we have
\(I=\int_{0}^{\pi / 4} \tan \theta \cdot \theta^{2} \cdot \sec ^{2} \theta d \theta=\int_{0}^{\pi / 4} \theta_{1}^{2}\left(\tan \theta \cdot \sec ^{2} \theta\right) d \theta\) 
\(=\left[\theta^{2} \cdot \frac{\tan ^{2} \theta}{2}\right]^{\pi / 4}-\int_{0}^{\pi / 4} 2 \theta \cdot \frac{\tan ^{2} \theta}{2} d \theta\) 
\(\left[\because \int \tan x \sec ^{2} x d x=\frac{\tan ^{2} x}{2}\right]\) 
\( =\frac{1}{2}\left[\frac{\pi^{2}}{16}\left(\tan \frac{\pi}{4}\right)^{2}-0\right]-\int_{0}^{\pi / 4} \theta \cdot\left(\sec ^{2} \theta-1\right) d \theta\ {\left[\because \sec ^{2} \theta-\tan ^{2} \theta=1\right]} \) 
\(\begin{array}{r} =\frac{1}{2}\left[\frac{\pi^{2}}{16}\left(\tan \frac{\pi}{4}\right)^{2}-0\right]-\int_{0}^{\pi / 4} \theta \cdot\left(\sec ^{2} \theta-1\right) d \theta\ {\left[\because \sec ^{2} \theta-\tan ^{2} \theta=1\right]} \end{array}\) 
\(=\frac{\pi^{2}}{32}-\left[\theta \cdot(\tan \theta-\theta)-\int 1 \cdot(\tan \theta-\theta) d \theta\right]_{0}^{\pi / 4}\) 
\(=\frac{\pi^{2}}{32}-\left[\theta \cdot \tan \theta-\theta^{2}\right]_{0}^{\pi / 4}+\int_{0}^{\pi / 4}(\tan \theta-\theta) d \theta\) 
\(=\frac{\pi^{2}}{32}-\left[\frac{\pi}{4} \tan \frac{\pi}{4}-\frac{\pi^{2}}{16}-0\right]+\left[\log |\sec \theta|-\frac{\theta^{2}}{2}\right]_{0}^{\pi / 4}\) 
\(=\frac{\pi^{2}}{32}-\frac{\pi}{4}+\frac{\pi^{2}}{16}+\log \sqrt{2}-\frac{\pi^{2}}{32}-0 \quad\left[\because \tan \left(\frac{\pi}{4}\right)=1\right]\) 
\(=\frac{\pi^{2}-4 \pi}{16}+\log \sqrt{2}\)