What is the integral from 0 to 3 of 1/((2-x)^2) ?

The integral from 0 to 3 of 1/((2-x)^2) is diverges

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integral from 0 to 3 of 1/((2-x)^2)

\int_{0}^{3} \frac{1}{( 2 - x ) ^{2}} d x

diverges

Solution steps

\int_{0}^{3} \frac{1}{( 2 - x ) ^{2}} d x

Find undefined (singularity) points:

x = 2

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}^{2} \frac{1}{( 2 - x ) ^{2}} d x + \int_{2}^{3} \frac{1}{( 2 - x ) ^{2}} d x

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{2}^{2} 1 + t^{3} d t

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

\int_{- 1}^{3} x^{2} - 1 d x

\int_{0}^{4} 3 x + 4 d x

\int_{0}^{1} t^{7} (1 + t^{8})^{7} d t

What is the integral from 0 to 3 of 1/((2-x)^2) ?
The integral from 0 to 3 of 1/((2-x)^2) is diverges