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

The integral from 1 to 2 of (x+1)/(sqrt(x)) is (4sqrt(2)-2)/3+2sqrt(2)-2

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

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

\frac{4 2 - 2}{3} + 22 - 2

Decimal

2.04737 \dots

Solution steps

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

Expand

\frac{x + 1}{x} : x + \frac{1}{x}

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

\int_{1}^{2} x d x = \frac{4 2 - 2}{3}

\int_{1}^{2} \frac{1}{x} d x = 22 - 2

= \frac{4 2 - 2}{3} + 22 - 2

What is the integral from 1 to 2 of (x+1)/(sqrt(x)) ?
The integral from 1 to 2 of (x+1)/(sqrt(x)) is (4sqrt(2)-2)/3+2sqrt(2)-2