What is the integral from 2 to 3 of 2/((3x-6)^2) ?

The integral from 2 to 3 of 2/((3x-6)^2) is diverges

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integral from 2 to 3 of 2/((3x-6)^2)

\int_{2}^{3} \frac{2}{( 3 x - 6 ) ^{2}} d x

diverges

Solution steps

\int_{2}^{3} \frac{2}{( 3 x - 6 ) ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{2}^{3} \frac{1}{( 3 x - 6 ) ^{2}} d x

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply the Power Rule

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

= 2 \cdot \frac{1}{3} [- \frac{1}{u}]_{0}^{3}

Simplify

2 \cdot \frac{1}{3} [- \frac{1}{u}]_{0}^{3} : \frac{2}{3} [- \frac{1}{u}]_{0}^{3}

= \frac{2}{3} [- \frac{1}{u}]_{0}^{3}

Compute the boundaries:

diverges

= diverges

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What is the integral from 2 to 3 of 2/((3x-6)^2) ?
The integral from 2 to 3 of 2/((3x-6)^2) is diverges