What is the integral from-6 to 10 of 1/(x+2) ?

The integral from-6 to 10 of 1/(x+2) is diverges

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integral from-6 to 10 of 1/(x+2)

\int_{- 6}^{10} \frac{1}{x + 2} d x

diverges

Solution steps

\int_{- 6}^{10} \frac{1}{x + 2} d x

Find undefined (singularity) points:

x = - 2

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_{- 6}^{- 2} \frac{1}{x + 2} d x + \int_{- 2}^{10} \frac{1}{x + 2} d x

\int_{- 6}^{- 2} \frac{1}{x + 2} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{0}^{1} x^{2} e^{- 10 x} d x

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\int_{0}^{\frac{π}{2}} sin^{6} (θ) d θ

\int_{0}^{\infty} 4 e^{8 x} d x

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

What is the integral from-6 to 10 of 1/(x+2) ?
The integral from-6 to 10 of 1/(x+2) is diverges