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

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

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

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

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

Solution steps

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

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

Add a constant to the solution

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

\int \frac{x ^{2}}{x ^{3} + 9} d x

\frac{d}{d x} ((a x^{2} + b x + c) e^{\frac{- x}{4}})

\frac{s ^{1}}{s ( s ^{2} + s + 1 )}

\frac{d}{d x} (9 x^{2} e^{x})

\frac{d}{d x} (x sinh (x) - cosh (x))

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