What is the integral of 1/(x^4-2x^3) ?

The integral of 1/(x^4-2x^3) is -1/8 ln|x|+1/(4x)+1/(4x^2)+1/8 ln|x-2|+C

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integral of 1/(x^4-2x^3)

\int \frac{1}{x ^{4} - 2 x ^{3}} d x

- \frac{1}{8} ln ∣ x ∣ + \frac{1}{4 x} + \frac{1}{4 x ^{2}} + \frac{1}{8} ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{1}{x ^{4} - 2 x ^{3}} d x

Take the partial fraction of

\frac{1}{x ^{4} - 2 x ^{3}} : - \frac{1}{8 x} - \frac{1}{4 x ^{2}} - \frac{1}{2 x ^{3}} + \frac{1}{8 ( x - 2 )}

= \int - \frac{1}{8 x} - \frac{1}{4 x ^{2}} - \frac{1}{2 x ^{3}} + \frac{1}{8 ( x - 2 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{1}{8 x} d x - \int \frac{1}{4 x ^{2}} d x - \int \frac{1}{2 x ^{3}} d x + \int \frac{1}{8 ( x - 2 )} d x

\int \frac{1}{8 x} d x = \frac{1}{8} ln ∣ x ∣

\int \frac{1}{4 x ^{2}} d x = - \frac{1}{4 x}

\int \frac{1}{2 x ^{3}} d x = - \frac{1}{4 x ^{2}}

\int \frac{1}{8 ( x - 2 )} d x = \frac{1}{8} ln ∣ x - 2 ∣

= - \frac{1}{8} ln ∣ x ∣ - (- \frac{1}{4 x}) - (- \frac{1}{4 x ^{2}}) + \frac{1}{8} ln ∣ x - 2 ∣

Simplify

= - \frac{1}{8} ln ∣ x ∣ + \frac{1}{4 x} + \frac{1}{4 x ^{2}} + \frac{1}{8} ln ∣ x - 2 ∣

Add a constant to the solution

= - \frac{1}{8} ln ∣ x ∣ + \frac{1}{4 x} + \frac{1}{4 x ^{2}} + \frac{1}{8} ln ∣ x - 2 ∣ + C

What is the integral of 1/(x^4-2x^3) ?
The integral of 1/(x^4-2x^3) is -1/8 ln|x|+1/(4x)+1/(4x^2)+1/8 ln|x-2|+C