What is the integral of int(x tan^-1 x)/(1+x^2)^(3/2)xtan1x(1+x2)32 ?

int(x tan^-1 x)/(1+x^2)^(3/2)xtan1x(1+x2)32

1 Answer
Jul 13, 2018

(1-tan^-1(x))/sqrt(x^2+1)+C1tan1(x)x2+1+C

Explanation:

I=int(xtan^-1(x))/(1+x^2)^(3/2)color(red)(dx)I=xtan1(x)(1+x2)32dx

Let x=tanthetax=tanθ, which implies that dx=sec^2thetad thetadx=sec2θdθ.

Then:

I=int(tanthetatan^-1(tantheta))/(1+tan^2theta)^(3/2)sec^2thetad thetaI=tanθtan1(tanθ)(1+tan2θ)32sec2θdθ

Remember that 1+tan^2theta=sec^2theta1+tan2θ=sec2θ:

I=int(thetatantheta)/sec^3thetasec^2thetad thetaI=θtanθsec3θsec2θdθ

I=intthetatantheta 1/secthetad thetaI=θtanθ1secθdθ

I=intthetasintheta/costhetacosthetad thetaI=θsinθcosθcosθdθ

I=intthetasinthetad thetaI=θsinθdθ

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 x=tantheta. Imagine the right triangle associated with this: we must have the leg opposite theta being x and the adjacent leg being 1. Then the hypotenuse is sqrt(x^2+1).

Thus costheta=1/sqrt(x^2+1) and sintheta=x/sqrt(x^2+1). Moreover, note that theta=tan^-1(x).

I=-tan^-1(x)/sqrt(x^2+1)+1/sqrt(x^2+1)+C

I=(1-tan^-1(x))/sqrt(x^2+1)+C