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

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

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

\int_{0}^{2} \frac{x ^{2} - 3}{x + 1} d x

- 2 ln (3)

Decimal Notation

- 2.19722 \dots

Solution steps

\int_{0}^{2} \frac{x ^{2} - 3}{x + 1} d x

Apply u-substitution

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

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

Expand

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

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

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

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

\int_{1}^{3} \frac{2}{u} d u = 2 ln (3)

= 4 - 4 - 2 ln (3)

Simplify

= - 2 ln (3)

\frac{d}{d x} (\frac{ln ( x )}{2 + ln ( x )})

x \to - 3 lim (2 x^{2})

\frac{d}{d x} (\frac{8}{x ^{2} - 4})

\frac{d}{d t} (t sin (2 t))

\frac{\partial}{\partial v} (v \cdot e^{u (v)})

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