How do you find the antiderivative of int (arctan(sqrtx)) dx(arctan(x))dx?

1 Answer
Nov 7, 2016

(x+1)arctan(sqrtx)-sqrtx+C(x+1)arctan(x)x+C

Explanation:

I=intarctan(sqrtx)dxI=arctan(x)dx

Integration by parts takes the form intudv=uv-intvduudv=uvvdu. So, for intarctan(sqrtx)dxarctan(x)dx, we should let u=arctan(sqrtx)u=arctan(x) and dv=dxdv=dx.

Differentiating uu gives (du)/dx=1/(1+(sqrtx)^2)*d/dxsqrtxdudx=11+(x)2ddxx or du=1/(2sqrtx(1+x))dxdu=12x(1+x)dx. From dv=dxdv=dx we integrate to show that v=xv=x. Thus:

I=uv-intvduI=uvvdu

color(white)(I)=xarctan(sqrtx)-intx/(2sqrtx(1+x))dxI=xarctan(x)x2x(1+x)dx

color(white)(I)=xarctan(sqrtx)-1/2intsqrtx/(1+x)dxI=xarctan(x)12x1+xdx

Letting t=sqrtxt=x implies that t^2=xt2=x and furthermore, that 2tdt=dx2tdt=dx through differentiating. Thus:

I=xarctan(sqrtx)-1/2intt/(1+t^2)(2tdt)I=xarctan(x)12t1+t2(2tdt)

color(white)I=xarctan(sqrtx)-intt^2/(1+t^2)dtI=xarctan(x)t21+t2dt

color(white)I=xarctan(sqrtx)-int(1+t^2-1)/(1+t^2)dtI=xarctan(x)1+t211+t2dt

color(white)I=xarctan(sqrtx)-int(1+t^2)/(1+t^2)dt+int1/(1+t^2)dtI=xarctan(x)1+t21+t2dt+11+t2dt

color(white)I=xarctan(sqrtx)-intdt+arctan(t)I=xarctan(x)dt+arctan(t)

color(white)I=xarctan(sqrtx)-t+arctan(t)I=xarctan(x)t+arctan(t)

Since t=sqrtxt=x:

I=xarctan(sqrtx)-sqrtx+arctan(sqrtx)I=xarctan(x)x+arctan(x)

color(white)I=(x+1)arctan(sqrtx)-sqrtx+CI=(x+1)arctan(x)x+C