How do you integrate $\frac{{tan}^{- 1} x}{1 + x^{2}}$ $tan^-1x/(1+x^2)$ ?

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How do you integrate $\frac{{tan}^{- 1} x}{1 + x^{2}}$ $tan^-1x/(1+x^2)$ ?

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Answer 1 · 2016-12-24T21:26:09

$\frac{{tan}^{- 1} x}{2} + C$ $tan^-1x/2 + C$

Explanation:

Let $u = {tan}^{- 1} x$ $u = tan^-1x$ . Then $d u = \frac{1}{x^{2} + 1} d x$ $du = 1/(x^2 + 1)dx$ and $d x = \frac{d u}{\frac{1}{x^{2} + 1}} = x^{2} + 1 d u$ $dx = (du)/(1/(x^2 + 1)) = x^2 + 1du$

Substitute:

$\Rightarrow \int \frac{u}{1 + x^{2}} \cdot 1 + x^{2} d u$ $=> intu/(1 + x^2) * 1 + x^2 du$

$\Rightarrow \int u d u$ $=> intu du$

Use the power rule $\int (x^{n}) = \frac{x^{n + 1}}{n + 1} + C$ $int(x^n) = x^(n + 1)/(n + 1) + C$ .

$\Rightarrow \frac{1}{2} u^{2} + C$ $=> 1/2u^2 + C$

$\Rightarrow \frac{1}{2} ({tan}^{- 1} x) + C$ $=> 1/2(tan^-1x) + C$

$\Rightarrow \frac{{tan}^{- 1} x}{2} + C$ $=> tan^-1x/2 + C$

Hopefully this helps!

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