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)

integral

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

\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

\frac{x}{( 1 + x ^{2} ) ^{2}}

x^{2} (e^{- x})^{2}

1 - x

2 t^{2}

\frac{1}{cos ^{3} ( x )}

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