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

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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

y^{^{'}} = \frac{( x - y sin ( 2 x ) )}{sin ^{2} ( x )}

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

3 e^{- 7 x}

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

\frac{4}{2 s ^{2} + 2 s + 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