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

1 Answer
maganbhai P.
Mar 6, 2018

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^2udurArrdx=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

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