What is the integral from 0 to 4 of x/(x^2-9) ?

The integral from 0 to 4 of x/(x^2-9) is diverges

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integral from 0 to 4 of x/(x^2-9)

\int_{0}^{4} \frac{x}{x ^{2} - 9} d x

diverges

Solution steps

\int_{0}^{4} \frac{x}{x ^{2} - 9} 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_{0}^{3} \frac{x}{x ^{2} - 9} d x + \int_{3}^{4} \frac{x}{x ^{2} - 9} d x

\int_{0}^{3} \frac{x}{x ^{2} - 9} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int \frac{5 x + 4}{x - 1} d x

\frac{d ^{2}}{d x ^{2}} (5 e^{x} cos (x))

x \frac{d y}{d x} + x y = 1 - y, y (1) = 0

(e^{x} + y) d x + (2 + x + y e^{y}) d y = 0, y (0) = 0

y = x^{3}, y = x^{2}, x = - 1

What is the integral from 0 to 4 of x/(x^2-9) ?
The integral from 0 to 4 of x/(x^2-9) is diverges