What is the integral from-1 to 1 of 9/((x+2)^2) ?

The integral from-1 to 1 of 9/((x+2)^2) is 6

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integral from-1 to 1 of 9/((x+2)^2)

\int_{- 1}^{1} \frac{9}{( x + 2 ) ^{2}} d x

6

Solution steps

\int_{- 1}^{1} \frac{9}{( x + 2 ) ^{2}} d x

Take the constant out:

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

= 9 \cdot \int_{- 1}^{1} \frac{1}{( x + 2 ) ^{2}} d x

Apply u-substitution

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

= 9 \cdot \int_{1}^{3} \frac{1}{u ^{2}} d u

Apply the Power Rule

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

= 9 [- \frac{1}{u}]_{1}^{3}

Compute the boundaries:

\frac{2}{3}

= 9 \cdot \frac{2}{3}

Simplify

= 6

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What is the integral from-1 to 1 of 9/((x+2)^2) ?
The integral from-1 to 1 of 9/((x+2)^2) is 6