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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{4} - 4 x

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

Add a constant to the solution

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

\int (sec (x) - cot (x))^{2} d x

\frac{x ^{9} + 7}{x ^{9}}

\frac{d}{d x} (1 0^{(x^{2} - 1)})

\int \frac{15 e ^{5 x}}{e ^{5 x} + 2} d x

f (x) = - 4 x^{2} + 7 x - 14, at x = - 2

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