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

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

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integral from-3 to 3 of 1/(x(x+1))

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

diverges

Solution steps

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

Find undefined (singularity) points:

x = 0, 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_{- 3}^{- 1} \frac{1}{x ( x + 1 )} d x + \int_{- 1}^{0} \frac{1}{x ( x + 1 )} d x + \int_{0}^{3} \frac{1}{x ( x + 1 )} d x

\int_{- 3}^{- 1} \frac{1}{x ( x + 1 )} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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

\int_{0}^{3} x (x^{2} + 2)^{2} d x

\int_{0}^{9} 8 - \frac{64 x}{9} d x

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

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

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