What is the integral from 1 to b of 1/((x+4)^2) ?

The integral from 1 to b of 1/((x+4)^2) is -1/(b+4)+1/5

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integral from 1 to b of 1/((x+4)^2)

\int_{1}^{b} \frac{1}{( x + 4 ) ^{2}} d x

- \frac{1}{b + 4} + \frac{1}{5}

Solution steps

\int_{1}^{b} \frac{1}{( x + 4 ) ^{2}} d x

Apply u-substitution

: \int_{5}^{b + 4} \frac{1}{u ^{2}} d u

= \int_{5}^{b + 4} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{5}^{b + 4}

= [- \frac{1}{u}]_{5}^{b + 4}

Compute the boundaries:

- \frac{1}{b + 4} + \frac{1}{5}

= - \frac{1}{b + 4} + \frac{1}{5}

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What is the integral from 1 to b of 1/((x+4)^2) ?
The integral from 1 to b of 1/((x+4)^2) is -1/(b+4)+1/5