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

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

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

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

\frac{1}{x + 2} + C

Solution steps

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

Take the constant out:

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

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

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

Simplify

= \frac{1}{x + 2}

Add a constant to the solution

= \frac{1}{x + 2} + C

x y^{^{'}} = 4 y

\int \frac{1}{49 e ^{- 4 x} + e ^{4 x}} d x

ln (1 + 4 x), 0

\frac{25}{s ^{2} + 6 s + 25}

\int arctan (8 t) d t