What is the integral from 0 to-4 of x/(16-2x^2) ?

The integral from 0 to-4 of x/(16-2x^2) is diverges

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integral from 0 to-4 of x/(16-2x^2)

\int_{0}^{- 4} \frac{x}{16 - 2 x ^{2}} d x

diverges

Solution steps

\int_{0}^{- 4} \frac{x}{16 - 2 x ^{2}} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 4}^{0} \frac{x}{16 - 2 x ^{2}} d x

Find undefined (singularity) points:

x = - 22

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_{- 4}^{- 22} \frac{x}{16 - 2 x ^{2}} d x + \int_{- 22}^{0} \frac{x}{16 - 2 x ^{2}} d x)

\int_{- 4}^{- 22} \frac{x}{16 - 2 x ^{2}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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What is the integral from 0 to-4 of x/(16-2x^2) ?
The integral from 0 to-4 of x/(16-2x^2) is diverges