What is $\int {tan}^{- 1} x d x$ $int tan^-1 x dx$ ?

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What is $\int {tan}^{- 1} x d x$ $int tan^-1 x dx$ ?

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Answer 1 · 2018-02-20T09:48:50

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

Explanation:

We want to solve

$I = \int {tan}^{- 1} (x) d x$ $I=inttan^-1(x)dx$

Use integration by parts / partial integration

$\int u d v = u v - \int v d u$ $intudv=uv-intvdu$

Let $u = {tan}^{- 1} (x)$ $u=tan^-1(x)$ and $d v = 1 d x$ $dv=1dx$

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

$I = {tan}^{- 1} (x) x - \int \frac{x}{x^{2} + 1} d x$ $I=tan^-1(x)x-intx/(x^2+1)dx$

Make a substitution $u = x^{2} + 1 \Rightarrow \frac{d u}{d x} = 2 x$ $u=x^2+1=>(du)/dx=2x$

$I = {tan}^{- 1} (x) x - \frac{1}{2} \int \frac{1}{u} d u$ $I=tan^-1(x)x-1/2int1/(u)du$

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

Substitute back $u = x^{2} + 1$ $u=x^2+1$

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

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What is $\int {tan}^{- 1} x d x$ $int tan^-1 x dx$ ?

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