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#