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

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

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

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

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

Decimal Notation

0.42463 \dots

Solution steps

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

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

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

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

\int_{0}^{1} 1 d x = 1

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

x \to - 8 lim (\frac{x ^{2} + 7}{x + 8})

(- 4, - 1), (3, 6)

\int x^{2} e^{x} d x

\int_{2}^{10} - 4 d x

\frac{1}{s ^{2} + π ^{2}}

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