What is the integral from 0 to 1 of 1/(x^4-1) ?

The integral from 0 to 1 of 1/(x^4-1) is diverges

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integral from 0 to 1 of 1/(x^4-1)

\int_{0}^{1} \frac{1}{x ^{4} - 1} d x

diverges

Solution steps

\int_{0}^{1} \frac{1}{x ^{4} - 1} d x

Take the partial fraction of

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

= \int_{0}^{1} - \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_{0}^{1} \frac{1}{2 ( x ^{2} + 1 )} d x - \int_{0}^{1} \frac{1}{4 ( x + 1 )} d x + \int_{0}^{1} \frac{1}{4 ( x - 1 )} d x

\int_{0}^{1} \frac{1}{2 ( x ^{2} + 1 )} d x = \frac{π}{8}

\int_{0}^{1} \frac{1}{4 ( x + 1 )} d x = \frac{1}{4} ln (2)

\int_{0}^{1} \frac{1}{4 ( x - 1 )} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{1}^{9} \frac{u - 2 u ^{2}}{u} d u

\int_{0}^{2} \frac{x}{x ^{2} - 1} d x

\int_{0}^{π} 1 + sin (2 x) d x

\int_{0}^{\frac{π}{2}} \frac{1}{1 - sin ( x )} d x

\int_{0}^{4} x - 1 d x

What is the integral from 0 to 1 of 1/(x^4-1) ?
The integral from 0 to 1 of 1/(x^4-1) is diverges