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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{3} ln ∣ u ∣

Substitute back

u = 1 - x^{3}

= - \frac{1}{3} ln 1 - x^{3}

Add a constant to the solution

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

y = 8 - x^{2}, y = x^{2}

(x^{2} + 5)^{3}

y^{^{''}} + 3 y^{^{'}} = 12

\int \frac{e ^{5 x}}{e ^{5 x} + 3} d x

x \frac{d y}{d x} - 3 y = 0

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