What is the integral from 0 to e of (x^2-9)/(x+3) ?

The integral from 0 to e of (x^2-9)/(x+3) is (e^2+6e)/2-6e

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integral from 0 to e of (x^2-9)/(x+3)

\int_{0}^{e} \frac{x ^{2} - 9}{x + 3} d x

\frac{e ^{2} + 6 e}{2} - 6 e

Decimal Notation

- 4.46031 \dots

Solution steps

\int_{0}^{e} \frac{x ^{2} - 9}{x + 3} d x

Expand

\frac{x ^{2} - 9}{x + 3} : \frac{x ^{2}}{x + 3} - \frac{9}{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}^{e} \frac{x ^{2}}{x + 3} d x - \int_{0}^{e} \frac{9}{x + 3} d x

\int_{0}^{e} \frac{x ^{2}}{x + 3} d x = \frac{e ^{2} + 6 e}{2} - 6 e + 9 (ln (e + 3) - ln (3))

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

= \frac{e ^{2} + 6 e}{2} - 6 e + 9 (ln (e + 3) - ln (3)) - 9 (ln (e + 3) - ln (3))

Simplify

= \frac{e ^{2} + 6 e}{2} - 6 e

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What is the integral from 0 to e of (x^2-9)/(x+3) ?
The integral from 0 to e of (x^2-9)/(x+3) is (e^2+6e)/2-6e