What is the integral from 0 to 6 of 2/((x+2)^2) ?

The integral from 0 to 6 of 2/((x+2)^2) is 3/4

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integral from 0 to 6 of 2/((x+2)^2)

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

\frac{3}{4}

Decimal

0.75

Solution steps

\int_{0}^{6} \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}^{6} \frac{1}{( x + 2 ) ^{2}} d x

Apply u-substitution

= 2 \cdot \int_{2}^{8} \frac{1}{u ^{2}} d u

Apply the Power Rule

= 2 [- \frac{1}{u}]_{2}^{8}

Compute the boundaries:

\frac{3}{8}

= 2 \cdot \frac{3}{8}

Simplify

= \frac{3}{4}

What is the integral from 0 to 6 of 2/((x+2)^2) ?
The integral from 0 to 6 of 2/((x+2)^2) is 3/4