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

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

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

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

4 (\frac{1}{x} - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣) + C

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

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

= 4 \cdot \int - \frac{1}{x ^{2}} - \frac{1}{2 ( x + 1 )} + \frac{1}{2 ( x - 1 )} 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

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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