Question

How do I find the integral `intarctan(4x)dx` ?

Answer 1

`I=x*tan^-1(4x)-1/4log|sqrt(1+16x^2)|+C`
`=x*tan^-1(4x)-1/8log|(1+16x^2)|+C`

Explanation:

`(1)I=inttan^-1(4x)dx`
Let, `tan^-1(4x)=urArr4x=tanurArr4dx=sec^2udu` `rArrdx=1/4sec^2udu`
`I=intu*1/4sec^2udu=1/4intu*sec^2udu`
Using Integration by Parts, `I=1/4[u*intsec^2udu-int(d/(du)(u)*intsec^2udu)du]=1/4[u*tanu-int1*tanudu]` `=1/4[u*tanu-log|secu|]+C` `=1/4[tan^-1(4x)*(4x)-log|sqrt(1+tan^2u|]+C` `=x*tan^-1(4x)-1/4log|sqrt(1+16x^2)|+C`
Second Method:
`(2)I=int1*tan^-1(4x)dx` `=tan^-1(4x)*x-int(1/(1+16x^2)*4)xdx`
`=x*tan^-1(4x)-1/8int(32x)/(1+16x^2)dx`
`=x*tan^-1(4x)-1/8log|1+16x^2|+C`