What is the integral from 4 to 7 of 1/((x-3)^2) ?

The integral from 4 to 7 of 1/((x-3)^2) is 3/4

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integral from 4 to 7 of 1/((x-3)^2)

\int_{4}^{7} \frac{1}{( x - 3 ) ^{2}} d x

\frac{3}{4}

Decimal

0.75

Solution steps

\int_{4}^{7} \frac{1}{( x - 3 ) ^{2}} d x

Apply u-substitution

= \int_{1}^{4} \frac{1}{u ^{2}} d u

Apply the Power Rule

= [- \frac{1}{u}]_{1}^{4}

Compute the boundaries:

\frac{3}{4}

= \frac{3}{4}

What is the integral from 4 to 7 of 1/((x-3)^2) ?
The integral from 4 to 7 of 1/((x-3)^2) is 3/4