What is the integral from 0 to 4 of 1/(x^2-2x-3) ?

The integral from 0 to 4 of 1/(x^2-2x-3) is diverges

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integral from 0 to 4 of 1/(x^2-2x-3)

\int_{0}^{4} \frac{1}{x ^{2} - 2 x - 3} d x

diverges

Solution steps

\int_{0}^{4} \frac{1}{x ^{2} - 2 x - 3} 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{1}{x ^{2} - 2 x - 3} d x + \int_{3}^{4} \frac{1}{x ^{2} - 2 x - 3} d x

\int_{0}^{3} \frac{1}{x ^{2} - 2 x - 3} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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

\frac{1}{g ^{2} ( x )}

10 x (x^{2} + 5)^{4}

y = 2 e^{3} t

\int (4 x (3 x - 2)) d x

What is the integral from 0 to 4 of 1/(x^2-2x-3) ?
The integral from 0 to 4 of 1/(x^2-2x-3) is diverges