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

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

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

\int_{- 2}^{2} \frac{1}{( x + 1 ) ^{4}} d x

diverges

Solution steps

\int_{- 2}^{2} \frac{1}{( x + 1 ) ^{4}} d x

Find undefined (singularity) points:

x = - 1

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_{- 2}^{- 1} \frac{1}{( x + 1 ) ^{4}} d x + \int_{- 1}^{2} \frac{1}{( x + 1 ) ^{4}} d x

\int_{- 2}^{- 1} \frac{1}{( x + 1 ) ^{4}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{d}{d x} (\frac{4 x}{5})

f^{^{'}} (x) = 12 x^{3} - 63 x^{2} - 72 x + 4

y = \frac{f ( x )}{x ^{6}}

y = a csc (2) (k x)

\int_{0}^{1} x - x^{k} d x

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