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

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

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

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

diverges

Solution steps

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{0}^{π} 2 + 2 cos (x) d x

\int_{\frac{π}{3}}^{\frac{π}{2}} cos (x) d x

\int_{2}^{4} x^{2} - 2 x d x

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

\int_{- 2}^{3} \frac{3}{4 x ^{2}} d x

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