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

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

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

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

\frac{1}{4} ln x^{4} - 2 + C

Solution steps

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

= \frac{1}{4} ln x^{4} - 2

Add a constant to the solution

= \frac{1}{4} ln x^{4} - 2 + C

f (x) = \frac{1}{x - 2}

\frac{\partial}{\partial x} (\frac{- 4 x ^{5}}{4 y ^{4}})

\frac{d y}{d x} y = x

- x^{2} + 3 x + 7

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