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

The integral from 0 to 1 of (3x-2)/(x+2) is 8ln(2)-8ln(3)+3

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

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

8 ln (2) - 8 ln (3) + 3

Decimal

- 0.24372 \dots

Solution steps

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

Expand

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

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

\int_{0}^{1} \frac{3 x}{x + 2} d x = 3 (1 - 2 (ln (3) - ln (2)))

\int_{0}^{1} \frac{2}{x + 2} d x = 2 (ln (3) - ln (2))

= 3 (1 - 2 (ln (3) - ln (2))) - 2 (ln (3) - ln (2))

Simplify

3 (1 - 2 (ln (3) - ln (2))) - 2 (ln (3) - ln (2)) : 8 ln (2) - 8 ln (3) + 3

= 8 ln (2) - 8 ln (3) + 3

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What is the integral from 0 to 1 of (3x-2)/(x+2) ?
The integral from 0 to 1 of (3x-2)/(x+2) is 8ln(2)-8ln(3)+3