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

1 Answer
May 27, 2018

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