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

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

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

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

diverges

Solution steps

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

Take the partial fraction of

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

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

n = 1 \sum \infty \frac{1}{n \cdot 3 ^{n}}

x \to - \infty lim (3 - x)

in v erse l a pl a ce \frac{4}{2 s ^{2} + 2 s + 1}

d er i v a t i v e y = e^{4} x^{2} + 1

\frac{\partial}{\partial y} (3 x^{2} y^{2} + 1)

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