What is the integral from 0 to 1 of 1/(x^3-5x^2) ?

The integral from 0 to 1 of 1/(x^3-5x^2) is diverges

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integral from 0 to 1 of 1/(x^3-5x^2)

\int_{0}^{1} \frac{1}{x ^{3} - 5 x ^{2}} d x

diverges

Solution steps

\int_{0}^{1} \frac{1}{x ^{3} - 5 x ^{2}} d x

Take the partial fraction of

\frac{1}{x ^{3} - 5 x ^{2}} : - \frac{1}{25 x} - \frac{1}{5 x ^{2}} + \frac{1}{25 ( x - 5 )}

= \int_{0}^{1} - \frac{1}{25 x} - \frac{1}{5 x ^{2}} + \frac{1}{25 ( x - 5 )} 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}{25 x} d x - \int_{0}^{1} \frac{1}{5 x ^{2}} d x + \int_{0}^{1} \frac{1}{25 ( x - 5 )} d x

\int_{0}^{1} \frac{1}{25 x} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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What is the integral from 0 to 1 of 1/(x^3-5x^2) ?
The integral from 0 to 1 of 1/(x^3-5x^2) is diverges