What is the integral of (x^3-1)/((4x^2-1)^2) ?

The integral of (x^3-1)/((4x^2-1)^2) is -3/32 ln|2x+1|+9/(64(2x+1))+5/32 ln|2x-1|+7/(64(2x-1))+C

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integral of (x^3-1)/((4x^2-1)^2)

\int \frac{x ^{3} - 1}{( 4 x ^{2} - 1 ) ^{2}} d x

- \frac{3}{32} ln ∣ 2 x + 1 ∣ + \frac{9}{64 ( 2 x + 1 )} + \frac{5}{32} ln ∣ 2 x - 1 ∣ + \frac{7}{64 ( 2 x - 1 )} + C

Solution steps

\int \frac{x ^{3} - 1}{( 4 x ^{2} - 1 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{3} - 1}{( 4 x ^{2} - 1 ) ^{2}} : - \frac{3}{16 ( 2 x + 1 )} - \frac{9}{32 ( 2 x + 1 ) ^{2}} + \frac{5}{16 ( 2 x - 1 )} - \frac{7}{32 ( 2 x - 1 ) ^{2}}

= \int - \frac{3}{16 ( 2 x + 1 )} - \frac{9}{32 ( 2 x + 1 ) ^{2}} + \frac{5}{16 ( 2 x - 1 )} - \frac{7}{32 ( 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{3}{16 ( 2 x + 1 )} d x - \int \frac{9}{32 ( 2 x + 1 ) ^{2}} d x + \int \frac{5}{16 ( 2 x - 1 )} d x - \int \frac{7}{32 ( 2 x - 1 ) ^{2}} d x

\int \frac{3}{16 ( 2 x + 1 )} d x = \frac{3}{32} ln ∣ 2 x + 1 ∣

\int \frac{9}{32 ( 2 x + 1 ) ^{2}} d x = - \frac{9}{64 ( 2 x + 1 )}

\int \frac{5}{16 ( 2 x - 1 )} d x = \frac{5}{32} ln ∣ 2 x - 1 ∣

\int \frac{7}{32 ( 2 x - 1 ) ^{2}} d x = - \frac{7}{64 ( 2 x - 1 )}

= - \frac{3}{32} ln ∣ 2 x + 1 ∣ - (- \frac{9}{64 ( 2 x + 1 )}) + \frac{5}{32} ln ∣ 2 x - 1 ∣ - (- \frac{7}{64 ( 2 x - 1 )})

Simplify

= - \frac{3}{32} ln ∣ 2 x + 1 ∣ + \frac{9}{64 ( 2 x + 1 )} + \frac{5}{32} ln ∣ 2 x - 1 ∣ + \frac{7}{64 ( 2 x - 1 )}

Add a constant to the solution

= - \frac{3}{32} ln ∣ 2 x + 1 ∣ + \frac{9}{64 ( 2 x + 1 )} + \frac{5}{32} ln ∣ 2 x - 1 ∣ + \frac{7}{64 ( 2 x - 1 )} + C

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What is the integral of (x^3-1)/((4x^2-1)^2) ?
The integral of (x^3-1)/((4x^2-1)^2) is -3/32 ln|2x+1|+9/(64(2x+1))+5/32 ln|2x-1|+7/(64(2x-1))+C