What is the integral of (9x^2)/(sqrt(x^3+1)) ?

The integral of (9x^2)/(sqrt(x^3+1)) is 6sqrt(x^3+1)+C

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integral of (9x^2)/(sqrt(x^3+1))

\int \frac{9 x ^{2}}{x ^{3} + 1} d x

6 x^{3} + 1 + C

Solution steps

\int \frac{9 x ^{2}}{x ^{3} + 1} d x

Take the constant out:

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

= 9 \cdot \int \frac{x ^{2}}{x ^{3} + 1} d x

Apply u-substitution

: \int \frac{1}{3 u} d u

= 9 \cdot \int \frac{1}{3 u} d u

Take the constant out:

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

= 9 \cdot \frac{1}{3} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= 9 \cdot \frac{1}{3} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = x^{3} + 1

= 9 \cdot \frac{1}{3} \cdot 2 (x^{3} + 1)^{\frac{1}{2}}

Simplify

9 \cdot \frac{1}{3} \cdot 2 (x^{3} + 1)^{\frac{1}{2}} : 6 x^{3} + 1

= 6 x^{3} + 1

Add a constant to the solution

= 6 x^{3} + 1 + C

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What is the integral of (9x^2)/(sqrt(x^3+1)) ?
The integral of (9x^2)/(sqrt(x^3+1)) is 6sqrt(x^3+1)+C