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

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

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

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

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

Solution steps

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

Factor

2 x^{2} - 1 : - (- 2 x^{2} + 1)

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

(- (- 2 x^{2} + 1))^{2} = (- 2 x^{2} + 1)^{2}

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

Apply u-substitution

: \int - \frac{1}{4 u ^{2}} d u

= \int - \frac{1}{4 u ^{2}} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{4} (- \frac{1}{u})

Substitute back

u = - 2 x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

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