What is the integral from 0 to 2 of ((x^2-2)/(x+1)) ?

The integral from 0 to 2 of ((x^2-2)/(x+1)) is -ln(3)

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integral from 0 to 2 of ((x^2-2)/(x+1))

\int_{0}^{2} (\frac{x ^{2} - 2}{x + 1}) d x

- ln (3)

Solution steps

\int_{0}^{2} \frac{x ^{2} - 2}{x + 1} d x

Apply u-substitution

: \int_{1}^{3} \frac{u ^{2} - 2 u - 1}{u} d u

= \int_{1}^{3} \frac{u ^{2} - 2 u - 1}{u} d u

Expand

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

= \int_{1}^{3} 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_{1}^{3} u d u - \int_{1}^{3} 2 d u - \int_{1}^{3} \frac{1}{u} d u

\int_{1}^{3} u d u = 4

\int_{1}^{3} 2 d u = 4

\int_{1}^{3} \frac{1}{u} d u = ln (3)

= 4 - 4 - ln (3)

Simplify

= - ln (3)

\int_{1}^{4} x^{- \frac{1}{2}} d x

\int_{0}^{4} (y - 2) (2 y + 1) d y

\int_{0}^{3} (3 x - 8) d x

\int_{3}^{4} x^{4} d x

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

What is the integral from 0 to 2 of ((x^2-2)/(x+1)) ?
The integral from 0 to 2 of ((x^2-2)/(x+1)) is -ln(3)