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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

(t^{2} + 16)^{2} y^{^{'}} + 2 t (t^{2} + 16) y = 4, y (- 2) = 10

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

\int 36 - 4 x^{2} d x

\int 3 x e^{- 3 x} d x

y = x^{2} - 2 x, y = 3 x + 6

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