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

The integral of 1/((4x^2-1)^2) is 1/8 ln|2x+1|-1/(8(2x+1))-1/8 ln|2x-1|-1/(8(2x-1))+C

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

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

\frac{1}{8} ln ∣ 2 x + 1 ∣ - \frac{1}{8 ( 2 x + 1 )} - \frac{1}{8} ln ∣ 2 x - 1 ∣ - \frac{1}{8 ( 2 x - 1 )} + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{4 ( 2 x + 1 )} d x = \frac{1}{8} ln ∣ 2 x + 1 ∣

\int \frac{1}{4 ( 2 x + 1 ) ^{2}} d x = - \frac{1}{8 ( 2 x + 1 )}

\int \frac{1}{4 ( 2 x - 1 )} d x = \frac{1}{8} ln ∣ 2 x - 1 ∣

\int \frac{1}{4 ( 2 x - 1 ) ^{2}} d x = - \frac{1}{8 ( 2 x - 1 )}

= \frac{1}{8} ln ∣ 2 x + 1 ∣ - \frac{1}{8 ( 2 x + 1 )} - \frac{1}{8} ln ∣ 2 x - 1 ∣ - \frac{1}{8 ( 2 x - 1 )}

Add a constant to the solution

= \frac{1}{8} ln ∣ 2 x + 1 ∣ - \frac{1}{8 ( 2 x + 1 )} - \frac{1}{8} ln ∣ 2 x - 1 ∣ - \frac{1}{8 ( 2 x - 1 )} + C

\int 2 x (x^{2} + 8)^{7} d x

\frac{\partial}{\partial x} ((y - x^{2})^{2} + x^{5})

g (x) = \frac{1}{( 1 + 2 x ) ^{3}}

1 - cos (x)

\int_{- π}^{π} π sin^{2} (x) d x

What is the integral of 1/((4x^2-1)^2) ?
The integral of 1/((4x^2-1)^2) is 1/8 ln|2x+1|-1/(8(2x+1))-1/8 ln|2x-1|-1/(8(2x-1))+C