What is the integral from 1 to 3 of (x^2-2)/(x+1) ?

The integral from 1 to 3 of (x^2-2)/(x+1) is -ln(2)+2

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integral from 1 to 3 of (x^2-2)/(x+1)

\int_{1}^{3} \frac{x ^{2} - 2}{x + 1} d x

- ln (2) + 2

Decimal

1.30685 \dots

Solution steps

\int_{1}^{3} \frac{x ^{2} - 2}{x + 1} d x

Apply u-substitution

= \int_{2}^{4} \frac{u ^{2} - 2 u - 1}{u} d u

Expand

\frac{u ^{2} - 2 u - 1}{u} : u - 2 - \frac{1}{u}

= \int_{2}^{4} u - 2 - \frac{1}{u} d u

Apply the Sum Rule:

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

= \int_{2}^{4} u d u - \int_{2}^{4} 2 d u - \int_{2}^{4} \frac{1}{u} d u

\int_{2}^{4} u d u = 6

\int_{2}^{4} 2 d u = 4

\int_{2}^{4} \frac{1}{u} d u = ln (2)

= 6 - 4 - ln (2)

Simplify

= - ln (2) + 2

What is the integral from 1 to 3 of (x^2-2)/(x+1) ?
The integral from 1 to 3 of (x^2-2)/(x+1) is -ln(2)+2