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

The integral from 1 to b of 1/((x+8)^2) is -1/(b+8)+1/9

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integral from 1 to b of 1/((x+8)^2)

\int_{1}^{b} \frac{1}{( x + 8 ) ^{2}} d x

- \frac{1}{b + 8} + \frac{1}{9}

Solution steps

\int_{1}^{b} \frac{1}{( x + 8 ) ^{2}} d x

Apply u-substitution

: \int_{9}^{b + 8} \frac{1}{u ^{2}} d u

= \int_{9}^{b + 8} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{9}^{b + 8}

= [- \frac{1}{u}]_{9}^{b + 8}

Compute the boundaries:

- \frac{1}{b + 8} + \frac{1}{9}

= - \frac{1}{b + 8} + \frac{1}{9}

\int \frac{e ^{x}}{3 + 16 e ^{2 x}} d x

\int \frac{( 1 )}{( u ^{2} - 1 )} d u

f (x) = 5 x^{3}, at x = 2

\frac{\partial}{\partial x} (x y + cos (x))

\frac{d}{d x} f (x) = (2 x - 6)^{2} (x^{2} + x + 1)^{2}

What is the integral from 1 to b of 1/((x+8)^2) ?
The integral from 1 to b of 1/((x+8)^2) is -1/(b+8)+1/9