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

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

Popular >

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

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

diverges

Solution steps

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

Expand

\frac{x ^{3} - 1}{x ^{2} - 1} : \frac{x ^{3}}{x ^{2} - 1} - \frac{1}{x ^{2} - 1}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{2}^{7} \frac{8 x}{x ^{2} + 1} d x

\int_{0}^{1} (x - 1)^{3} d x

\int_{0}^{x} x cos (x) d x

\int_{0}^{1} x 56 x^{2} + 25 d x

\int_{- \frac{5 π}{6}}^{0} sin (x) d x

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