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

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

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

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

\frac{1}{2}

Decimal

0.5

Solution steps

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

Simplify

\frac{x ^{2} - 1}{x + 1} : x - 1

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

\int_{1}^{2} x d x = \frac{3}{2}

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

= \frac{3}{2} - 1

Simplify

\frac{3}{2} - 1 : \frac{1}{2}

= \frac{1}{2}

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