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Answer: $\frac{1}{\sqrt{1+x^{2}}}\left ( x-\tan ^{-1}x \right )+c$

Hint:  $x=\tan \theta$&$\int uvdx=u\int vdx-\int \left ( \frac{d}{dx}u\int vdx \right )dx$

Given: $\int \frac{x\tan ^{-1}x}{\left ( 1+x^{2} \right )^{\frac{3}{2}}}dx$

Solution:

$I= \frac{x\tan ^{-1}x}{\left ( 1+x^{2} \right )^{\frac{3}{2}}}dx$

Put $x=\tan \Theta \Rightarrow dx=\sec ^{2}\Theta \, d\: \Theta$

$\begin{gathered} I=\int \frac{\tan \theta \tan ^{-1}(\tan \theta) \sec ^{2} \theta d \theta}{\left(\sqrt{1+\tan ^{2} \theta}\right)^{\frac{3}{2}}} \\ =\int \frac{\tan \sec ^{2} \theta}{\sqrt{\sec ^{2} \theta} \sec ^{3} \theta} \\ =\int\left(\frac{\theta \tan \theta}{\sec \theta}\right) d \theta \end{gathered}$

\begin{aligned} &I=\int \theta \frac{\sin \theta}{\cos \theta} \times \cos \theta d \theta \\ &I=\theta \int \sin \theta d \theta-\int\left[\frac{d}{d \theta} \theta \int \sin \theta d \theta\right] d \theta \\ &=-\theta \cos \theta+\int(\cos \theta) d \theta \\ &=-\theta \cos \theta+\sin \theta+c \\ &=-\tan ^{-1} x\left(\frac{1}{\sqrt{1+x^{2}}}\right)+\frac{x}{\sqrt{1+x^{2}}}+c \\ &=\frac{1}{\sqrt{1+x^{2}}}\left(x-\tan ^{-1} x\right)+c \end{aligned}

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