What is the integral from-1 to 1 of 1/(x(x+4)) ?

The integral from-1 to 1 of 1/(x(x+4)) is diverges

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integral from-1 to 1 of 1/(x(x+4))

\int_{- 1}^{1} \frac{1}{x ( x + 4 )} d x

diverges

Solution steps

\int_{- 1}^{1} \frac{1}{x ( x + 4 )} d x

Find undefined (singularity) points:

x = 0

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}^{0} \frac{1}{x ( x + 4 )} d x + \int_{0}^{1} \frac{1}{x ( x + 4 )} d x

\int_{- 1}^{0} \frac{1}{x ( x + 4 )} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{d}{d x} (8 x - x^{2})

f (x) = 3 x^{7}

\frac{\partial}{\partial x} (\frac{3 x - t}{4 s + 5 t})

\frac{d}{d t} (4 t e^{- t})

f (x) = \frac{x ^{2}}{x ^{2} - 16}

What is the integral from-1 to 1 of 1/(x(x+4)) ?
The integral from-1 to 1 of 1/(x(x+4)) is diverges