Please help.. Integration?

Question

x^2 [sqrt[x^3=11]]dx

Answer 1

I assume you meant to type a + instead of an = sign.

Integral (x^2 * sqrt(x^3 + 11))dx

I'm going to rearrange those terms to show something to you.

Integral (sqrt(x^3 + 11) x^2 dx )

What we have to do is solve this one by substitution.
Let u = x^3 + 11
Then du = 3x^2 dx, and

(1/3) du = x^2 dx

NOTE: Look at the way I rearranged the integral. See the spot where I put x^2 dx, and see how this nicely equals (1/3) du according to our substitution? That's EXACTLY how we replace it. Thus, our integral becomes:

Integral (sqrt(u) (1/3)du )

It's always a good idea to pull out constants from integrals. This is a step we can do safely. Better to get them out of the way instead of having to deal with them by solving the integral. In our case, the constant is 1/3.

(1/3) * Integral (sqrt(u) du)

Note that sqrt(u) can be written as u^(1/2), and whenever we take integrals of powers, the formula for the integral of x^n is
(x^(n+1))/(n+1)

So our answer is then

(1/3) * [u^(1/2+1)/[1/2 + 1] + C

OR

(1/3) [u^(3/2) / [3/2] ] + C
(1/3) (2/3) [u^(3/2)] + C
(2/9) [u^(3/2)] + C

Replacing u = x^3 + 11, we get

(2/9) [ (x^3 + 11)^(3/2) ] + C

Answer 2

Try u = x^3/3. Then du =3x^2. The problem then becomes

Integral [sqrt(3u - 11) du]

This one can be solved by setting v = 3u - 11, dv = 3du. Get

Integral [(1/3)sqrt(v)dv]

This is essentially taking the integral of a square root. After doing this, substitute back and get the original integral.

Answer 3

put x^3+11=t
3x^2dx=dt
the integral reduces to
1/3rt t dt
1/3t^3/2/3/2
=2/9(x^3+11)^3/2+C