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

The integral from 1 to b of 1/((x+7)^2) is -1/(b+7)+1/8

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

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

- \frac{1}{b + 7} + \frac{1}{8}

Solution steps

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

Apply u-substitution

: \int_{8}^{b + 7} \frac{1}{u ^{2}} d u

= \int_{8}^{b + 7} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{8}^{b + 7}

= [- \frac{1}{u}]_{8}^{b + 7}

Compute the boundaries:

- \frac{1}{b + 7} + \frac{1}{8}

= - \frac{1}{b + 7} + \frac{1}{8}

\frac{d}{d x} (arccos (\frac{x}{a}))

y = (9 x^{2} + 4 x + 4)^{5}

3 - 2 e^{- 4 t}

\int \frac{2 x ^{2} + 5}{x ^{2} + 1} d x

\frac{d}{d y} (\frac{z}{x ^{2} + y ^{2} + z ^{2}})

What is the integral from 1 to b of 1/((x+7)^2) ?
The integral from 1 to b of 1/((x+7)^2) is -1/(b+7)+1/8