\(log\ |{x^2+2\over\sqrt{x^2+1}}|+c\)

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

\(tan\ x-cot\ x-3x+c\)

\({2\over3}sin^{-1}\sqrt{x^3\over a^3}+c\)

Let \(I=\int \frac{(x+5)}{3 x^{2}+13 x-10} d x\) 
\(=\int \frac{(x+5)}{(x+5)(3 x-2)} d x \)
\(=\int \frac{d x}{3 x-2}=\frac{1}{3} \log |3 x-2|+C \) 

Write the given integral as \(\int \frac{x^{2} \cdot x}{\left(x^{2}\right)^{2}+3 x^{2}+2} d x\) 
Put \(x^{2}=t\) then above integral reduces to 
\(\frac{1}{2} \int \frac{t}{t^{2}+3 t+2} d t\) 
\(\log \left|\frac{x^{2}+2}{\sqrt{x^{2}+1}}\right|+C\)

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

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

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
\(I=\int_{0}^{\pi / 4} \tan \theta \cdot \theta^{2} \cdot \sec ^{2} \theta d \theta=\int_{0}^{\pi / 4} \theta_{1}^{2}\left(\tan \theta \cdot \sec ^{2} \theta\right) d \theta\) 
\(=\left[\theta^{2} \cdot \frac{\tan ^{2} \theta}{2}\right]^{\pi / 4}-\int_{0}^{\pi / 4} 2 \theta \cdot \frac{\tan ^{2} \theta}{2} d \theta\) 
\(\left[\because \int \tan x \sec ^{2} x d x=\frac{\tan ^{2} x}{2}\right]\) 
\( =\frac{1}{2}\left[\frac{\pi^{2}}{16}\left(\tan \frac{\pi}{4}\right)^{2}-0\right]-\int_{0}^{\pi / 4} \theta \cdot\left(\sec ^{2} \theta-1\right) d \theta\ {\left[\because \sec ^{2} \theta-\tan ^{2} \theta=1\right]} \) 
\(\begin{array}{r} =\frac{1}{2}\left[\frac{\pi^{2}}{16}\left(\tan \frac{\pi}{4}\right)^{2}-0\right]-\int_{0}^{\pi / 4} \theta \cdot\left(\sec ^{2} \theta-1\right) d \theta\ {\left[\because \sec ^{2} \theta-\tan ^{2} \theta=1\right]} \end{array}\) 
\(=\frac{\pi^{2}}{32}-\left[\theta \cdot(\tan \theta-\theta)-\int 1 \cdot(\tan \theta-\theta) d \theta\right]_{0}^{\pi / 4}\) 
\(=\frac{\pi^{2}}{32}-\left[\theta \cdot \tan \theta-\theta^{2}\right]_{0}^{\pi / 4}+\int_{0}^{\pi / 4}(\tan \theta-\theta) d \theta\) 
\(=\frac{\pi^{2}}{32}-\left[\frac{\pi}{4} \tan \frac{\pi}{4}-\frac{\pi^{2}}{16}-0\right]+\left[\log |\sec \theta|-\frac{\theta^{2}}{2}\right]_{0}^{\pi / 4}\) 
\(=\frac{\pi^{2}}{32}-\frac{\pi}{4}+\frac{\pi^{2}}{16}+\log \sqrt{2}-\frac{\pi^{2}}{32}-0 \quad\left[\because \tan \left(\frac{\pi}{4}\right)=1\right]\) 
\(=\frac{\pi^{2}-4 \pi}{16}+\log \sqrt{2}\)

Let \(I=\int_{\pi / 3}^{\pi / 2} \frac{\sqrt{1+\cos x} d x}{(1-\cos x)^{5 / 2}}=\int_{\pi / 3}^{\pi / 2} \frac{\sqrt{1-\cos ^{2} x}}{(1-\cos x)^{\frac{5}{2}+\frac{1}{2}}} d x\) 
[multiplying numerator and denominator by \(\sqrt{1-\cos x}\) ]
\(=\int_{\pi / 3}^{\pi / 2} \frac{\sin x}{(1-\cos x)^{3}} d x\) 
(ii) Put 1 - cos x = t and simplify it. = \(\frac{3}{2}\)