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

The integral of 1/((8+x)^2) is -1/(8+x)+C

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integral of 1/((8+x)^2)

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

- \frac{1}{8 + x} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{2}} d u

= \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = 8 + x

= - \frac{1}{8 + x}

Add a constant to the solution

= - \frac{1}{8 + x} + C

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