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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{u}

= - 2 (- \frac{1}{u})

Substitute back

u = x + 1

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

Simplify

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

= \frac{2}{x + 1}

Add a constant to the solution

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

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\frac{d y}{d x} = 15 x y

\frac{\partial}{\partial x} (e^{- 2 x} cos (θ))