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

The integral of (3x^2+1)/((x^2-1)^3) is 1/(4(x+1)^2)-1/(4(x-1)^2)+C

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

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

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

Solution steps

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

Take the partial fraction of

\frac{3 x ^{2} + 1}{( x ^{2} - 1 ) ^{3}} : - \frac{1}{2 ( x + 1 ) ^{3}} + \frac{1}{2 ( x - 1 ) ^{3}}

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

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

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

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

Simplify

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

Add a constant to the solution

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

8 sin (8 x - 1)

\frac{d}{d x} (e^{x} (2 x + 3)^{3})

\frac{d ^{2}}{d x ^{2}} (e^{\frac{3}{x}})

\int \frac{9 x}{x ^{2} + 1} d x

f (x) = 6 x, at x = 1

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