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

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

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

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

diverges

Solution steps

\int_{0}^{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_{0}^{1} \frac{1}{( x - 1 ) ^{4}} d x + \int_{1}^{2} \frac{1}{( x - 1 ) ^{4}} d x

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{0}^{1} (- t + 1) e^{- s t} d t

\int_{0}^{\frac{1}{2}} 9 - 36 x^{2} d x

\int_{- 16}^{16} \frac{1}{256 + x ^{2}} d x

\int_{1}^{x} x y d y

\int_{0}^{1} (1 + x^{3}) d x

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