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

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

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

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

- \frac{1}{b + 6} + \frac{1}{7}

Solution steps

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

Apply u-substitution

: \int_{7}^{b + 6} \frac{1}{u ^{2}} d u

= \int_{7}^{b + 6} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{7}^{b + 6}

= [- \frac{1}{u}]_{7}^{b + 6}

Compute the boundaries:

- \frac{1}{b + 6} + \frac{1}{7}

= - \frac{1}{b + 6} + \frac{1}{7}

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