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

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

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integral from 0 to e+1 of (x+1)/(x^2-1)

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

diverges

Solution steps

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

Find undefined (singularity) points:

x = 1

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{1}^{2} (2 x + 5) d x

\int_{0}^{π} sin (x + n x) d x

\int_{0}^{2} x + 2 - x d x

\int_{2}^{4} \frac{x - 2}{x ^{2} - 5 x + 4} d x

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

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