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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

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

Add a constant to the solution

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

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