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

The integral of (9x^2)/(1-x^3) is -3ln|1-x^3|+C

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integral of (9x^2)/(1-x^3)

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

- 3 ln 1 - x^{3} + C

Solution steps

\int \frac{9 x ^{2}}{1 - x ^{3}} 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}}{1 - x ^{3}} 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 (- \frac{1}{3} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 9 (- \frac{1}{3} ln ∣ u ∣)

Substitute back

u = 1 - x^{3}

= 9 (- \frac{1}{3} ln 1 - x^{3})

Simplify

9 (- \frac{1}{3} ln 1 - x^{3}) : - 3 ln 1 - x^{3}

= - 3 ln 1 - x^{3}

Add a constant to the solution

= - 3 ln 1 - x^{3} + C

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What is the integral of (9x^2)/(1-x^3) ?
The integral of (9x^2)/(1-x^3) is -3ln|1-x^3|+C