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

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

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

\int_{- 3}^{3} \frac{1}{( x - 1 ) ^{4}} d x

diverges

Solution steps

\int_{- 3}^{3} \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_{- 3}^{1} \frac{1}{( x - 1 ) ^{4}} d x + \int_{1}^{3} \frac{1}{( x - 1 ) ^{4}} d x

\int_{- 3}^{1} \frac{1}{( x - 1 ) ^{4}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{x - 2 t}^{x + 2 t} sin^{2} (x) d x

\int_{- 3}^{3} (- 5 x^{2} - 6 x)^{3} d x

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

\int_{0}^{1} (2 e^{\frac{x}{3 + 4}} + 5) d x

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

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