Integration of Tan Inverse x

Here you will learn proof of integration of tan inverse x or arctan x and examples based on it.

Let’s begin –

Integration of Tan Inverse x

The integration of tan inverse x or arctan x is $xtan^{-1}x$ – $1\over 2$ $log |1 + x^2|$ + C

Where C is the integration constant.

i.e. $\int$ $tan^{-1}x$ = $xtan^{-1}x$ – $1\over 2$ $log |1 + x^2|$ + C

Proof : 

We have, I = $\int$ $tan^{-1}x$ dx

Let $tan^{-1}x$ = t,

Then, x = tan t

$\implies$ dx = d(tan t) = $sec^2 t$ dt

$\therefore$ I = $\int$ $tan^{-1}x$ dx

$\implies$ I = $\int$ t $sec^2 t$ dt

By using integration by parts formula,

I = t tan t – $\int$ 1. (tan t) dt

I = t tan t + log |cos t| + C

Since tan t = x $\implies$ cost = $1\over \sqrt{1 + tan^2 t}$ = $1\over \sqrt{1 + x^2}$

Now, Put t = $tan^{-1}x$ here,

$\implies$ I = x $tan^{-1}x$ + $log |{1\over \sqrt{ 1+ x^2}}|$ + C

Hence, $\int$ $tan^{-1}x$ dx = $xtan^{-1}x$ – $1\over 2$ $log |1 + x^2|$ + C

Example : Evaluate $\int$ $x tan^{-1} x$ dx

Solution : We have,

I = $\int$  $x tan^{-1} x$ dx

By using integration by parts formula,

I = $tan^{-1} x$ $x^2\over 2$ – $\int$ $1\over 1 + x^2$ $\times$ $x^2\over 2$ dx

I = $tan^{-1} x$ $x^2\over 2$ – $1\over 2$ $\int$ $x^2 + 1 – 1\over 1 + x^2$dx

= $x^2\over 2$ $tan^{-1} x$ – $1\over 2$ $\int$  1 – $1\over 1 + x^2$dx

$\implies$ I = $x^2\over 2$ $tan^{-1} x$ – $1\over 2$ (x  – $tan^{-1} x$) + C

$\implies$ I = $x^2\over 2$ $tan^{-1} x$ – $x\over 2$ + $tan^{-1} x\over 2$ + C

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Related Questions

What is the Differentiation of tan inverse x ?

What is the Integration of tan x ?

What is the integration of tan inverse root x ?