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

The integral from 0 to 3 of (3x-2)/(x+1) is 9-10ln(2)

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

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

9 - 10 ln (2)

Decimal Notation

2.06852 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{4} \frac{3 u - 5}{u} d u

= \int_{1}^{4} \frac{3 u - 5}{u} d u

Expand

\frac{3 u - 5}{u} : 3 - \frac{5}{u}

= \int_{1}^{4} 3 - \frac{5}{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}^{4} 3 d u - \int_{1}^{4} \frac{5}{u} d u

\int_{1}^{4} 3 d u = 9

\int_{1}^{4} \frac{5}{u} d u = 10 ln (2)

= 9 - 10 ln (2)

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What is the integral from 0 to 3 of (3x-2)/(x+1) ?
The integral from 0 to 3 of (3x-2)/(x+1) is 9-10ln(2)