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

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

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integral of (1+x)^{-2}

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

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

Solution steps

\int (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}

Add a constant to the solution

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

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