What is the integral of (x^2)/(sqrt(x^3+9)) ?

The integral of (x^2)/(sqrt(x^3+9)) is 2/3 sqrt(x^3+9)+C

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integral of (x^2)/(sqrt(x^3+9))

\int \frac{x ^{2}}{x ^{3} + 9} d x

\frac{2}{3} x^{3} + 9 + C

Solution steps

\int \frac{x ^{2}}{x ^{3} + 9} d x

Apply u-substitution

= \int \frac{2}{3} d u

Integral of a constant:

\int a d x = a x

= \frac{2}{3} u

Substitute back

u = x^{3} + 9

= \frac{2}{3} x^{3} + 9

Add a constant to the solution

= \frac{2}{3} x^{3} + 9 + C

What is the integral of (x^2)/(sqrt(x^3+9)) ?
The integral of (x^2)/(sqrt(x^3+9)) is 2/3 sqrt(x^3+9)+C