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

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

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

integral

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

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

Solution steps

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

Take the constant out:

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

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

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

Simplify

= \frac{1}{1 + x}

Add a constant to the solution

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

ln^{5} (x^{2})

\frac{x ^{3}}{x ^{2}}

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

\frac{1}{( 5 t + 2 )}

1 + e^{v}

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