What is the integral from-1 to 1 of (18)/((x+2)^2) ?

The integral from-1 to 1 of (18)/((x+2)^2) is 12

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integral from-1 to 1 of (18)/((x+2)^2)

\int_{- 1}^{1} \frac{18}{( x + 2 ) ^{2}} d x

12

Solution steps

\int_{- 1}^{1} \frac{18}{( x + 2 ) ^{2}} d x

Take the constant out:

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

= 18 \cdot \int_{- 1}^{1} \frac{1}{( x + 2 ) ^{2}} d x

Apply u-substitution

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

= 18 \cdot \int_{1}^{3} \frac{1}{u ^{2}} d u

Apply the Power Rule

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

= 18 [- \frac{1}{u}]_{1}^{3}

Compute the boundaries:

\frac{2}{3}

= 18 \cdot \frac{2}{3}

Simplify

= 12

\int_{0}^{3} - 12 x d x

\int_{- 2}^{0} 3 x^{3} - 12 x d x

\int_{0}^{4} (x^{2} + 6 x) d x

\int_{- 0.5}^{0} b + 2 b x d x

\int_{- 1}^{1} (x^{5} - 4 x^{2} + x - 1) d x

What is the integral from-1 to 1 of (18)/((x+2)^2) ?
The integral from-1 to 1 of (18)/((x+2)^2) is 12