Question #24569

1 Answer
Aug 7, 2016

(2sqrt(x+a)(x-2a))/3+C

Explanation:

We have:

intx/sqrt(x+a)dx

If we want to use substitution, a good bet would be to let u=x+a. This also would mean that du=dx. The only issue left is that we've accounted for both dx and sqrt(x+a) in the denominator, and still have the x in the numerator left over.

To deal with this, use the original substitution u=x+a to write x as x=u-a. Then, we have:

intx/sqrt(x+a)dx=int(u-a)/sqrtudu

At this point, split the integral through subtraction:

int(u-a)/sqrtudu=intu/sqrtudu-inta/sqrtudu

Rewrite using powers:

intu/sqrtudu-inta/sqrtudu=intu^(1/2)du-aintu^(-1/2)du

Integrate both using the power rule for integrals:

intu^(1/2)du-aintu^(-1/2)du=u^(3/2)/(3/2)-a(u^(1/2)/(1/2))+C

=(2u^(3/2))/3-2au^(1/2)+C

=(2sqrtu(u-3a))/3+C

Since u=x+a:

=(2sqrt(x+a)(x+a-3a))/3+C

=(2sqrt(x+a)(x-2a))/3+C