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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

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

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

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

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

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

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

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

Add a constant to the solution

= 3 (- \frac{1}{2} arctan (x) + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x - 1 ∣) + C

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