What is the integral from-5 to-2 of 1/((x+3)^2) ?

The integral from-5 to-2 of 1/((x+3)^2) is diverges

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integral from-5 to-2 of 1/((x+3)^2)

\int_{- 5}^{- 2} \frac{1}{( x + 3 ) ^{2}} d x

diverges

Solution steps

\int_{- 5}^{- 2} \frac{1}{( x + 3 ) ^{2}} d x

Find undefined (singularity) points:

x = - 3

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_{- 5}^{- 3} \frac{1}{( x + 3 ) ^{2}} d x + \int_{- 3}^{- 2} \frac{1}{( x + 3 ) ^{2}} d x

\int_{- 5}^{- 3} \frac{1}{( x + 3 ) ^{2}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{0}^{5} (- 2 x^{2} + 10 x) d x

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\int_{- 2}^{6} \frac{1}{( x - 6 ) ^{\frac{2}{3}}} d x

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

What is the integral from-5 to-2 of 1/((x+3)^2) ?
The integral from-5 to-2 of 1/((x+3)^2) is diverges