What is the integral of (9x^2)/(x^3+8) ?

The integral of (9x^2)/(x^3+8) is 3ln|x^3+8|+C

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integral of (9x^2)/(x^3+8)

\int \frac{9 x ^{2}}{x ^{3} + 8} d x

3 ln x^{3} + 8 + C

Solution steps

\int \frac{9 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

= 9 \cdot \int \frac{x ^{2}}{x ^{3} + 8} 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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 9 \cdot \frac{1}{3} ln ∣ u ∣

Substitute back

u = x^{3} + 8

= 9 \cdot \frac{1}{3} ln x^{3} + 8

Simplify

9 \cdot \frac{1}{3} ln x^{3} + 8 : 3 ln x^{3} + 8

= 3 ln x^{3} + 8

Add a constant to the solution

= 3 ln x^{3} + 8 + C

\int \frac{( x ^{2} - 8 )}{8 x} d x

\int_{0}^{2} (x^{3} - 6 x^{2} + 8 x) d x

\int_{0}^{1} 6 cos (x^{2}) d x

\frac{\partial}{\partial x} (4 x^{3} - 4 y)

\frac{d y}{d x} + \frac{3 y}{x} = 6 x^{2}

What is the integral of (9x^2)/(x^3+8) ?
The integral of (9x^2)/(x^3+8) is 3ln|x^3+8|+C