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

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

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

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

- ln (5) + 2 ln (2) + 1

Decimal Notation

0.77685 \dots

Solution steps

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

\frac{x + 2}{x + 3} = - \frac{1}{x + 3} + 1

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

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

\int_{1}^{2} \frac{1}{x + 3} d x = ln (5) - 2 ln (2)

\int_{1}^{2} 1 d x = 1

= - (ln (5) - 2 ln (2)) + 1

Simplify

= - ln (5) + 2 ln (2) + 1

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