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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

\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 )}

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

Add a constant to the solution

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

\int_{2.3}^{2.75} 2 d x

y^{^{''}} + y^{^{'}} + y = 0.5 x

x^{2} y^{^{'}} = x y + x^{2} e^{\frac{y}{x}}

\frac{d}{d x} (3 x + 2 x ln (x))

\frac{\partial}{\partial y} (e^{- 2 x} cos (2 y))

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