What is the integral from-3 to 4 of 1/(x+2) ?

The integral from-3 to 4 of 1/(x+2) is diverges

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integral from-3 to 4 of 1/(x+2)

\int_{- 3}^{4} \frac{1}{x + 2} d x

diverges

Solution steps

\int_{- 3}^{4} \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_{- 3}^{- 2} \frac{1}{x + 2} d x + \int_{- 2}^{4} \frac{1}{x + 2} d x

\int_{- 3}^{- 2} \frac{1}{x + 2} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{0}^{x} (t - 1)^{2} (t - 3)^{3} d t

\int_{0}^{0.2} \frac{x ^{6}}{1 + x ^{7}} d x

\int_{0}^{t} cos^{2} (x) d x

\int_{0}^{4} x - 3 d x

\int_{1}^{3} (- x^{3} + 4 x^{2} - 3) d x

What is the integral from-3 to 4 of 1/(x+2) ?
The integral from-3 to 4 of 1/(x+2) is diverges