Question #24569
1 Answer
Explanation:
We have:
intx/sqrt(x+a)dx
If we want to use substitution, a good bet would be to let
To deal with this, use the original substitution
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
=(2sqrt(x+a)(x+a-3a))/3+C
=(2sqrt(x+a)(x-2a))/3+C