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

1 Answer
Mar 6, 2018

Answer:

#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#