What is the integral from 0 to 2 of (x^2-6+x)/(x+3) ?

The integral from 0 to 2 of (x^2-6+x)/(x+3) is -2

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integral from 0 to 2 of (x^2-6+x)/(x+3)

\int_{0}^{2} \frac{x ^{2} - 6 + x}{x + 3} d x

- 2

Solution steps

\int_{0}^{2} \frac{x ^{2} - 6 + x}{x + 3} d x

Expand

\frac{x ^{2} - 6 + x}{x + 3} : \frac{x ^{2}}{x + 3} - \frac{6}{x + 3} + \frac{x}{x + 3}

Apply the Sum Rule:

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

= \int_{0}^{2} \frac{x ^{2}}{x + 3} d x - \int_{0}^{2} \frac{6}{x + 3} d x + \int_{0}^{2} \frac{x}{x + 3} d x

\int_{0}^{2} \frac{x ^{2}}{x + 3} d x = 9 ln (5) - 4 - 9 ln (3)

\int_{0}^{2} \frac{6}{x + 3} d x = 6 (ln (5) - ln (3))

\int_{0}^{2} \frac{x}{x + 3} d x = 2 - 3 (ln (5) - ln (3))

= 9 ln (5) - 4 - 9 ln (3) - 6 (ln (5) - ln (3)) + 2 - 3 (ln (5) - ln (3))

Simplify

9 ln (5) - 4 - 9 ln (3) - 6 (ln (5) - ln (3)) + 2 - 3 (ln (5) - ln (3)) : - 2

= - 2

What is the integral from 0 to 2 of (x^2-6+x)/(x+3) ?
The integral from 0 to 2 of (x^2-6+x)/(x+3) is -2