What is the integral from 0 to 5 of (2x-1)/(x+1) ?

The integral from 0 to 5 of (2x-1)/(x+1) is 10-3ln(6)

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integral from 0 to 5 of (2x-1)/(x+1)

\int_{0}^{5} \frac{2 x - 1}{x + 1} d x

10 - 3 ln (6)

Decimal Notation

4.62472 \dots

Solution steps

\int_{0}^{5} \frac{2 x - 1}{x + 1} d x

Apply u-substitution

: \int_{1}^{6} \frac{2 u - 3}{u} d u

= \int_{1}^{6} \frac{2 u - 3}{u} d u

Expand

\frac{2 u - 3}{u} : 2 - \frac{3}{u}

= \int_{1}^{6} 2 - \frac{3}{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}^{6} 2 d u - \int_{1}^{6} \frac{3}{u} d u

\int_{1}^{6} 2 d u = 10

\int_{1}^{6} \frac{3}{u} d u = 3 ln (6)

= 10 - 3 ln (6)

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What is the integral from 0 to 5 of (2x-1)/(x+1) ?
The integral from 0 to 5 of (2x-1)/(x+1) is 10-3ln(6)