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

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

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

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

diverges

Solution steps

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

Apply u-substitution

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

Apply the Power Rule

= [- \frac{1}{u}]_{0}^{2}

Compute the boundaries:

diverges

= diverges

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

\int_{- \infty}^{0} (x + 1) e^{x} d x

\int_{- \infty}^{\frac{2}{a}} x e^{a x} d x

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

\int_{0}^{2} 6 e^{3 x} d x

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