How do you find the integral of $tanˉ¹(2x) dx$ ?

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Answer 1 · 2018-07-14T14:18:38

The answer is $=xarctan(2x)-1/4ln(4x^2+1)+C$

Perform a substitution

$u=2x$ , $=>$ , $du=2dx$

The integral is

$I=intarctan(2x)dx$

$=1/2intarctanudu$

$intfg'dx=fg-intf'gdx$

Here,

$f=arctanu$ , $=>$ , $f'=1/(u^2+1)$

$g'=1$ , $=>$ , $g=u$

Therefore,

$I=1/2u*arctanu-1/2int(udu)/(u^2+1)$

$=1/2u*arctanu-1/4ln(u^2+1)$

$=1/2*2xarctan(2x)-1/4ln(4x^2+1)+C$

$=xarctan(2x)-1/4ln(4x^2+1)+C$

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