Question

Question #24569

Answer 1

`(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`