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

The integral from 1 to 2 of 1-(1+x)^{-2} is 5/6

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

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

\frac{5}{6}

Decimal

0.83333 \dots

Solution steps

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

Apply exponent rule:

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

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

\int_{1}^{2} 1 d x = 1

\int_{1}^{2} \frac{1}{( 1 + x ) ^{2}} d x = \frac{1}{6}

= 1 - \frac{1}{6}

Simplify

1 - \frac{1}{6} : \frac{5}{6}

= \frac{5}{6}

What is the integral from 1 to 2 of 1-(1+x)^{-2} ?
The integral from 1 to 2 of 1-(1+x)^{-2} is 5/6