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

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

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

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

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

Solution steps

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

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d x = x

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

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

Simplify

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

Add a constant to the solution

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

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