What is the integral of (x^2)/(4-x^3) ?

The integral of (x^2)/(4-x^3) is -1/3 ln|4-x^3|+C

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integral of (x^2)/(4-x^3)

\int \frac{x ^{2}}{4 - x ^{3}} d x

- \frac{1}{3} ln 4 - x^{3} + C

Solution steps

\int \frac{x ^{2}}{4 - 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 = 4 - x^{3}

= - \frac{1}{3} ln 4 - x^{3}

Add a constant to the solution

= - \frac{1}{3} ln 4 - x^{3} + C

x = 1 \sum \infty \frac{x}{x ^{2} + 1}

\int_{0}^{7} w^{2} ln (w) d w

\int cot^{3} (t) d t

y = - 3 cos (x), at x = \frac{- π}{4}

3 x + 1, x + 1

What is the integral of (x^2)/(4-x^3) ?
The integral of (x^2)/(4-x^3) is -1/3 ln|4-x^3|+C