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

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

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

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

- \frac{1}{x ^{2} - 1} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u ^{2}} d u

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

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{u}

= 2 \cdot \frac{1}{2} (- \frac{1}{u})

Substitute back

u = x^{2} - 1

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

Simplify

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

= - \frac{1}{x ^{2} - 1}

Add a constant to the solution

= - \frac{1}{x ^{2} - 1} + C

\frac{d y}{d x} = 3^{e^{2 x}} y^{2}

\frac{9}{8 x ^{4}}

x \to 1 lim (\frac{8}{1 - x})

2 x^{^{''}} + 4 x^{^{'}} + 8 x = 0

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

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