What is the integral of #int(x tan^-1 x)/(1+x^2)^(3/2)# ?
#int(x tan^-1 x)/(1+x^2)^(3/2)#
1 Answer
Explanation:
#I=int(xtan^-1(x))/(1+x^2)^(3/2)color(red)(dx)#
Let
Then:
#I=int(tanthetatan^-1(tantheta))/(1+tan^2theta)^(3/2)sec^2thetad theta#
Remember that
#I=int(thetatantheta)/sec^3thetasec^2thetad theta#
#I=intthetatantheta 1/secthetad theta#
#I=intthetasintheta/costhetacosthetad theta#
#I=intthetasinthetad theta#
This looks primed for integration by parts. Let:
#{(u=theta,du=d theta),(dv=sinthetad theta,v=-costheta):}#
Then:
#I=uv-intvdu#
#I=-thetacostheta+intcosthetad theta#
#I=-thetacostheta+sintheta+C#
Remember that
Thus
#I=-tan^-1(x)/sqrt(x^2+1)+1/sqrt(x^2+1)+C#
#I=(1-tan^-1(x))/sqrt(x^2+1)+C#