Question

Evaluate the integrals : 3x^2 sqr(x^3+1)dx?

Answer 1

`I=2/3(x^3+1)^(3/2)+C`

Explanation:

We want to solve

`I=int3x^2sqrt(x^3+1)dx`

Make a substitution `color(blue)(u=x^3+1=>du=3x^2dx`

`I=intsqrt(u)du=intu^(1/2)du`

By the power rule for integration `color(red)(intx^ndx=x^(n+1)/(n+1)+C`

`I=u^(1/2+1)/(1/2+1)+C=u^(3/2)/(3/2)+C=2/3u^(3/2)+C`

Substitute back `color(blue)(u=x^3+1`

`I=2/3(x^3+1)^(3/2)+C`