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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

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

= 2 \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

= 2 (- \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 )}

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

Add a constant to the solution

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

x \to - \infty lim (x - (x))

y d x - x d y = 0

\int_{2}^{5} x d x

n = 1 \sum \infty ln (1)

\frac{d}{d x} ((x^{2} + x)^{6})

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