What is the integral of (20)/((x+2)^2) ?

The integral of (20)/((x+2)^2) is -(20)/(x+2)+C

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integral of (20)/((x+2)^2)

\int \frac{20}{( x + 2 ) ^{2}} d x

- \frac{20}{x + 2} + C

Solution steps

\int \frac{20}{( x + 2 ) ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 20 \cdot \int \frac{1}{( x + 2 ) ^{2}} d x

Apply u-substitution

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

= 20 \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= 20 (- \frac{1}{u})

Substitute back

u = x + 2

= 20 (- \frac{1}{x + 2})

Simplify

20 (- \frac{1}{x + 2}) : - \frac{20}{x + 2}

= - \frac{20}{x + 2}

Add a constant to the solution

= - \frac{20}{x + 2} + C

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What is the integral of (20)/((x+2)^2) ?
The integral of (20)/((x+2)^2) is -(20)/(x+2)+C