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

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

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

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

- \frac{1}{2} arctan (x) - \frac{x}{4 ( 1 + x ^{2} )} - \frac{1}{8} ln ∣ x + 1 ∣ + \frac{1}{8} ln ∣ x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{( x ^{2} + 1 ) ^{2} ( x ^{2} - 1 )} : - \frac{1}{4 ( x ^{2} + 1 )} - \frac{1}{2 ( x ^{2} + 1 ) ^{2}} - \frac{1}{8 ( x + 1 )} + \frac{1}{8 ( x - 1 )}

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

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

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

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

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

= - \frac{1}{4} arctan (x) - \frac{1}{4} (arctan (x) + \frac{x}{1 + x ^{2}}) - \frac{1}{8} ln ∣ x + 1 ∣ + \frac{1}{8} ln ∣ x - 1 ∣

Simplify

- \frac{1}{4} arctan (x) - \frac{1}{4} (arctan (x) + \frac{x}{1 + x ^{2}}) - \frac{1}{8} ln ∣ x + 1 ∣ + \frac{1}{8} ln ∣ x - 1 ∣ : - \frac{1}{2} arctan (x) - \frac{x}{4 ( 1 + x ^{2} )} - \frac{1}{8} ln ∣ x + 1 ∣ + \frac{1}{8} ln ∣ x - 1 ∣

= - \frac{1}{2} arctan (x) - \frac{x}{4 ( 1 + x ^{2} )} - \frac{1}{8} ln ∣ x + 1 ∣ + \frac{1}{8} ln ∣ x - 1 ∣

Add a constant to the solution

= - \frac{1}{2} arctan (x) - \frac{x}{4 ( 1 + x ^{2} )} - \frac{1}{8} ln ∣ x + 1 ∣ + \frac{1}{8} ln ∣ x - 1 ∣ + C

3 y^{^{''}} + y^{^{'}} + 2 y = 0

\int \frac{4 x ^{2}}{x ^{3} + 8} d x

\frac{\partial}{\partial x} ((1 - x - 4 y)^{\frac{1}{2}})

\frac{\partial}{\partial x} (5 e^{x^{5} y})

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

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