What is the integral from 4 to 6 of 2/((x-3)^2) ?

The integral from 4 to 6 of 2/((x-3)^2) is 4/3

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integral from 4 to 6 of 2/((x-3)^2)

\int_{4}^{6} \frac{2}{( x - 3 ) ^{2}} d x

\frac{4}{3}

Decimal

1.33333 \dots

Solution steps

\int_{4}^{6} \frac{2}{( x - 3 ) ^{2}} d x

Take the constant out:

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

= 2 \cdot \int_{4}^{6} \frac{1}{( x - 3 ) ^{2}} d x

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{2}{3}

= 2 \cdot \frac{2}{3}

Simplify

= \frac{4}{3}

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What is the integral from 4 to 6 of 2/((x-3)^2) ?
The integral from 4 to 6 of 2/((x-3)^2) is 4/3