What is the integral of (x^3-1)/(x^4-4x) ?

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

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integral of (x^3-1)/(x^4-4x)

\int \frac{x ^{3} - 1}{x ^{4} - 4 x} d x

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

Solution steps

\int \frac{x ^{3} - 1}{x ^{4} - 4 x} 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} - 4 x

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (\frac{1}{\frac{1}{x} + \frac{1}{y}})

\frac{\partial}{\partial x} (4 arctan (x y))

\int 12 d x

x \to 0 lim (\frac{x}{tan ( 2 x )})

n = 8 \sum \infty \frac{n + 3}{n !}

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