How do you find the integral intx^3/(sqrt(x^2+9))dx ?

1 Answer
Aug 29, 2014

=(x^2+9)^(3/2)/3-9sqrt(x^2+9)+c, where c is a constant

Explanation :

=intx^3/sqrt(x^2+9)dx

Using Integration by Substitution,

Let's assume x^2+9=t^2, then

2xdx=2tdt, =>xdx=tdt

=int((t^2-9)t)/tdt

=intt^2dt-int9dt

=t^3/3-9t+c, where c is a constant

Substituting t back,

=(x^2+9)^(3/2)/3-9sqrt(x^2+9)+c, where c is a constant