Evaluate the following integrals: (x ^ -1 x) (1+x^2)^ 3 2 dx

Question

Evaluate the following integrals:
$\int\frac{(\text{x}\tan^{-1}\text{x})}{(1+\text{x}^2)^{\frac{3}{2}}}\text{dx}$

✓

Answer

Let $\text{I}=\int\frac{\text{x}\tan^{-1}\text{x}}{(1+\text{x}^2)^{\frac{3}{2}}}\text{dx}$
putting $\text{x}=\tan\theta$
$\Rightarrow\text{dx}=\sec^2 \theta\text{d}\theta$
$\&\theta\tan^{-1}\text{x}$
$\therefore\text{I}=\int\frac{(\tan\theta).\theta\sec^2\theta\text{d}\theta}{\big(1+\tan^2\theta\big)^{\frac{3}{2}}}$
$=\int\frac{\theta.\tan\theta\sec^2\theta\text{d}\theta}{(\sec^2\theta)^{\frac{3}{2}}}$
$=\int\frac{\theta\tan\theta.\sec^2\theta\text{d}\theta}{\sec^3\theta}$
$=\int\frac{\theta.\tan\theta}{\sec\theta}\text{d}\theta$
$=\int\theta.\sin\theta\text{d}\theta$
$=\theta\int\sin\theta\text{d}\theta-\int\big\{\frac{\text{d}}{\text{d}\theta}(\theta)\int\sin\text{d}\theta\big\}\text{d}\theta$
$=\theta(-\cos\theta)-\int1.(-\cos\theta)\text{d}\theta$
$=-\theta\cos\theta+\sin\theta+\text{C}$
$=\frac{-\theta}{\sec\theta}+\frac{1}{\text{cosec}\theta}+\text{C}$
$=\frac{-\theta}{\sqrt{1+\tan^2\theta}}+\frac{1}{\sqrt{1+\cot^2}\theta}+\text{C}$
$=\frac{-\theta}{\sqrt{1+\tan^2\theta}}+\frac{\tan\theta}{\sqrt{\tan^2\theta+1}}+\text{C}$
$=\frac{-\tan^{-1}\text{x}}{\sqrt{1+\text{x}^2}}+\frac{\text{x}}{\sqrt{\text{x}^2+1}}+\text{C}$

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