What is the integral of (3x^2)/(sqrt(x^3+8)) ?

The integral of (3x^2)/(sqrt(x^3+8)) is 2sqrt(x^3+8)+C

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integral of (3x^2)/(sqrt(x^3+8))

\int \frac{3 x ^{2}}{x ^{3} + 8} d x

2 x^{3} + 8 + C

Solution steps

\int \frac{3 x ^{2}}{x ^{3} + 8} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int \frac{x ^{2}}{x ^{3} + 8} d x

Apply u-substitution

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

Integral of a constant:

\int a d x = a x

= 3 \cdot \frac{2}{3} u

Substitute back

u = x^{3} + 8

= 3 \cdot \frac{2}{3} x^{3} + 8

Simplify

3 \cdot \frac{2}{3} x^{3} + 8 : 2 x^{3} + 8

= 2 x^{3} + 8

Add a constant to the solution

= 2 x^{3} + 8 + C

What is the integral of (3x^2)/(sqrt(x^3+8)) ?
The integral of (3x^2)/(sqrt(x^3+8)) is 2sqrt(x^3+8)+C