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

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

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

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

ln (3) + 2

Decimal Notation

3.09861 \dots

Solution steps

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

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

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

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

\int_{0}^{2} 1 d x = 2

= ln (3) + 2

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y = 35 x - 16 x^{2}, at x = 1

h \to 1 lim (\frac{∣ h - 2 ∣ - 2}{h})

8 + 5 x + 7 x e^{x}

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