What is the integral from 2 to 5 of 1/((x-2)^2) ?

The integral from 2 to 5 of 1/((x-2)^2) is diverges

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integral from 2 to 5 of 1/((x-2)^2)

\int_{2}^{5} \frac{1}{( x - 2 ) ^{2}} d x

diverges

Solution steps

\int_{2}^{5} \frac{1}{( x - 2 ) ^{2}} d x

Apply u-substitution

: \int_{0}^{3} \frac{1}{u ^{2}} d u

= \int_{0}^{3} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{0}^{3}

= [- \frac{1}{u}]_{0}^{3}

Compute the boundaries:

diverges

= diverges

\int_{- π}^{π} sin (5 x) cos (3 x) d x

\int_{- 1}^{1} x (7 - x^{2}) d x

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\int_{1}^{5} 1 + (\frac{1}{x})^{2} d x

\int_{1}^{5} 2 x ln (x^{3}) d x

What is the integral from 2 to 5 of 1/((x-2)^2) ?
The integral from 2 to 5 of 1/((x-2)^2) is diverges