How do you evaluate the integral int arctansqrtx?

1 Answer
mason m
Jan 15, 2017

intarctan(sqrtx)dx=(x+1)arctan(sqrtx)-sqrtx+C

Explanation:

I=intarctan(sqrtx)dx

Let t=sqrtx. Finding dx in a usable form is simpler if we first write that t^2=x, which then implies that 2tdt=dx. This way we can plug in dx into the integral straight away. These substitutions yield:

I=intarctan(t)(2tdt)=int2tarctan(t)dt

Now use integration by parts. Let:

{(u=arctan(t)" "=>" "du=1/(t^2+1)dt),(dv=2tdt" "=>" "v=t^2):}

Then:

I=t^2arctan(t)-intt^2/(t^2+1)dt

Rewriting the integrand as (t^2+1-1)/(t^2+1)=(t^2+1)/(t^2+1)-1/(t^2+1)=1-1/(t^2+1) we see that

I=t^2arctan(t)-(intdt-int1/(t^2+1)dt)

Both of which are common integrals:

I=t^2arctan(t)-t+arctan(t)+C

Returning to x from t=sqrtx:

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

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

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