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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

= 4 \cdot \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

= 4 (\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 ∣

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

Simplify

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

\frac{d}{d x} ((x^{2} + 3)^{5})

f (x) = (sin (x))^{2}

x^{2} + 1, 0, 2

\frac{\partial}{\partial x} (f (x, y) (x y))

x \to 0 - lim (\frac{4 - x}{x})

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