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

The integral of (x^2)/(8-x^3) is -1/3 ln|8-x^3|+C

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integral of (x^2)/(8-x^3)

\int \frac{x ^{2}}{8 - x ^{3}} d x

- \frac{1}{3} ln 8 - x^{3} + C

Solution steps

\int \frac{x ^{2}}{8 - 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 = 8 - x^{3}

= - \frac{1}{3} ln 8 - x^{3}

Add a constant to the solution

= - \frac{1}{3} ln 8 - x^{3} + C

\frac{d}{d t} (\frac{1}{2 + t})

\frac{d}{d z} (\frac{z}{z + \frac{a}{z}})

y^{^{''}} + 2 y^{^{'}} = 8 x

(2 x + 1) \frac{d y}{d x} = 4 x y

\int \frac{sec ( x ) + cos ( x )}{3 cos ( x )} d x

What is the integral of (x^2)/(8-x^3) ?
The integral of (x^2)/(8-x^3) is -1/3 ln|8-x^3|+C