What is the integral from 4 to 6 of 1/((x-4)^2) ?

The integral from 4 to 6 of 1/((x-4)^2) is diverges

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integral from 4 to 6 of 1/((x-4)^2)

\int_{4}^{6} \frac{1}{( x - 4 ) ^{2}} d x

diverges

Solution steps

\int_{4}^{6} \frac{1}{( x - 4 ) ^{2}} d x

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

diverges

= diverges

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\int_{2}^{4.5} \frac{x}{18} d x

\int_{1}^{e^{2}} \frac{( in ( x ) ) ^{2}}{x} d x

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

What is the integral from 4 to 6 of 1/((x-4)^2) ?
The integral from 4 to 6 of 1/((x-4)^2) is diverges