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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{2} ( u - 2 )} d u

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

Take the partial fraction of

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

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

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

\int \frac{1}{4 u} d u = \frac{1}{4} ln ∣ u ∣

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

\int \frac{1}{4 ( u - 2 )} d u = \frac{1}{4} ln ∣ u - 2 ∣

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

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

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