What is the integral from 0 to 4 of x/(x+9) ?

The integral from 0 to 4 of x/(x+9) is 4-9(ln(13)-2ln(3))

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integral from 0 to 4 of x/(x+9)

\int_{0}^{4} \frac{x}{x + 9} d x

4 - 9 (ln (13) - 2 ln (3))

Decimal

0.69047 \dots

Solution steps

\int_{0}^{4} \frac{x}{x + 9} d x

Apply u-substitution

= \int_{9}^{13} \frac{u - 9}{u} d u

Expand

\frac{u - 9}{u} : 1 - \frac{9}{u}

= \int_{9}^{13} 1 - \frac{9}{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_{9}^{13} 1 d u - \int_{9}^{13} \frac{9}{u} d u

\int_{9}^{13} 1 d u = 4

\int_{9}^{13} \frac{9}{u} d u = 9 (ln (13) - 2 ln (3))

= 4 - 9 (ln (13) - 2 ln (3))

What is the integral from 0 to 4 of x/(x+9) ?
The integral from 0 to 4 of x/(x+9) is 4-9(ln(13)-2ln(3))