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

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

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integral of 1/(1-x^4)

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

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

Solution steps

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

Take the partial fraction of

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

= \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

= \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 ∣

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

Add a constant to the solution

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

7 e^{u}

\frac{d}{d x} (7^{3^{x^{2}}})

4 + 7 cos (5 t)

\int \frac{6}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{d y}{( 5 - 3 y ) ^{2}} = \frac{d x}{2 ( 4 - x ) ^{2}}

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