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

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

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

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

ln ∣ x + 1 ∣ - \frac{1}{x + 1} - ln ∣ x - 1 ∣ - \frac{1}{x - 1} + C

Solution steps

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

(x^{2} - 1)^{2} = x^{4} - 2 x^{2} + 1

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

Take the partial fraction of

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

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

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

\int \frac{1}{4 ( x + 1 )} d x = \frac{1}{4} ln ∣ x + 1 ∣

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

\int \frac{1}{4 ( x - 1 )} d x = \frac{1}{4} ln ∣ x - 1 ∣

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

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

Simplify

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

= ln ∣ x + 1 ∣ - \frac{1}{x + 1} - ln ∣ x - 1 ∣ - \frac{1}{x - 1}

Add a constant to the solution

= ln ∣ x + 1 ∣ - \frac{1}{x + 1} - ln ∣ x - 1 ∣ - \frac{1}{x - 1} + C

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