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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

= 3 (- \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4 ( x - 1 )}) + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4 ( x + 1 )} - \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4 ( x - 1 )}

Simplify

3 (- \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4 ( x - 1 )}) + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4 ( x + 1 )} - \frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{4 ( x - 1 )} : - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ - \frac{1}{x + 1} - \frac{1}{x - 1}

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

Add a constant to the solution

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

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