What is the integral from 1 to 10 of 1/(x^2-4x-32) ?

The integral from 1 to 10 of 1/(x^2-4x-32) is diverges

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integral from 1 to 10 of 1/(x^2-4x-32)

\int_{1}^{10} \frac{1}{x ^{2} - 4 x - 32} d x

diverges

Solution steps

\int_{1}^{10} \frac{1}{x ^{2} - 4 x - 32} d x

Find undefined (singularity) points:

x = 8

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_{1}^{8} \frac{1}{x ^{2} - 4 x - 32} d x + \int_{8}^{10} \frac{1}{x ^{2} - 4 x - 32} d x

\int_{1}^{8} \frac{1}{x ^{2} - 4 x - 32} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

g (x) = \frac{1}{5 x ^{2} + 4 x}

\frac{ln ( x )}{3 x}, \frac{( ln ( x ) ) ^{2}}{3 x}

y^{^{'}} - \frac{3}{x} y = \frac{y ^{5}}{x ^{6}}

\int_{3}^{\infty} e^{- \frac{x}{2}} d x

\int \frac{1}{( e ^{x} + e ^{- x} ) ^{2}} d x

What is the integral from 1 to 10 of 1/(x^2-4x-32) ?
The integral from 1 to 10 of 1/(x^2-4x-32) is diverges