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

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

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

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

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

Solution steps

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

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

Simplify

= \frac{1}{x + 1}

Add a constant to the solution

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

(x^{2} + 1) \frac{d y}{d x} + 3 x (y - 1) = 0

\frac{d}{d x} (\frac{x}{tan ( x )})

\frac{2 e ^{- 2 s}}{s ( s + 2 ) ( s + 3 ) ^{2}}

x \to a lim (2 f (x) - 3 g (x))

\frac{d}{d x} ((x^{2} + a^{2})^{e^{x^{2}}})