What is the integral of (x^3)/(8-x^4) ?

The integral of (x^3)/(8-x^4) is -1/4 ln|8-x^4|+C

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integral of (x^3)/(8-x^4)

\int \frac{x ^{3}}{8 - x ^{4}} d x

- \frac{1}{4} ln 8 - x^{4} + C

Solution steps

\int \frac{x ^{3}}{8 - x ^{4}} d x

Apply u-substitution

: \int - \frac{1}{4 u} d u

= \int - \frac{1}{4 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{4} ln ∣ u ∣

Substitute back

u = 8 - x^{4}

= - \frac{1}{4} ln 8 - x^{4}

Add a constant to the solution

= - \frac{1}{4} ln 8 - x^{4} + C

\frac{d}{d x} (arcsinh (x))

\frac{d y}{d x} y^{3} = 7 x^{3} + 2 x

\int (\frac{6}{4 - x ^{2}}) d x

\frac{d}{d y} (e^{x^{2} + y^{2}} sin (x y))

x^{4} - 3 x

What is the integral of (x^3)/(8-x^4) ?
The integral of (x^3)/(8-x^4) is -1/4 ln|8-x^4|+C