What is the integral from 0 to 7 of 2/((x+2)^2) ?

The integral from 0 to 7 of 2/((x+2)^2) is 7/9

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integral from 0 to 7 of 2/((x+2)^2)

\int_{0}^{7} \frac{2}{( x + 2 ) ^{2}} d x

\frac{7}{9}

Decimal Notation

0.77777 \dots

Solution steps

\int_{0}^{7} \frac{2}{( x + 2 ) ^{2}} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{7} \frac{1}{( x + 2 ) ^{2}} d x

Apply u-substitution

: \int_{2}^{9} \frac{1}{u ^{2}} d u

= 2 \cdot \int_{2}^{9} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{2}^{9}

= 2 [- \frac{1}{u}]_{2}^{9}

Compute the boundaries:

\frac{7}{18}

= 2 \cdot \frac{7}{18}

Simplify

= \frac{7}{9}

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What is the integral from 0 to 7 of 2/((x+2)^2) ?
The integral from 0 to 7 of 2/((x+2)^2) is 7/9