What is the integral from 0 to 6 of 1/((2x+7)^2) ?

The integral from 0 to 6 of 1/((2x+7)^2) is 6/133

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integral from 0 to 6 of 1/((2x+7)^2)

\int_{0}^{6} \frac{1}{( 2 x + 7 ) ^{2}} d x

\frac{6}{133}

Decimal Notation

0.04511 \dots

Solution steps

\int_{0}^{6} \frac{1}{( 2 x + 7 ) ^{2}} d x

Apply u-substitution

: \int_{7}^{19} \frac{1}{2 u ^{2}} d u

= \int_{7}^{19} \frac{1}{2 u ^{2}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{7}^{19} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{7}^{19}

= \frac{1}{2} [- \frac{1}{u}]_{7}^{19}

Compute the boundaries:

\frac{12}{133}

= \frac{1}{2} \cdot \frac{12}{133}

Simplify

= \frac{6}{133}

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What is the integral from 0 to 6 of 1/((2x+7)^2) ?
The integral from 0 to 6 of 1/((2x+7)^2) is 6/133