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

The integral from 0 to 2 of (3x+1)/(sqrt(x)) is 6sqrt(2)

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integral from 0 to 2 of (3x+1)/(sqrt(x))

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

62

Decimal

8.48528 \dots

Solution steps

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

Expand

\frac{3 x + 1}{x} : 3 x + \frac{1}{x}

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

\int_{0}^{2} 3 x d x = 42

\int_{0}^{2} \frac{1}{x} d x = 22

= 42 + 22

Simplify

= 62

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What is the integral from 0 to 2 of (3x+1)/(sqrt(x)) ?
The integral from 0 to 2 of (3x+1)/(sqrt(x)) is 6sqrt(2)