What is the integral from 1 to 9 of (x+1)/(sqrt(x)) ?

The integral from 1 to 9 of (x+1)/(sqrt(x)) is 64/3

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integral from 1 to 9 of (x+1)/(sqrt(x))

\int_{1}^{9} \frac{x + 1}{x} d x

\frac{64}{3}

Decimal

21.33333 \dots

Solution steps

\int_{1}^{9} \frac{x + 1}{x} d x

Expand

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

= \int_{1}^{9} 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}^{9} x d x + \int_{1}^{9} \frac{1}{x} d x

\int_{1}^{9} x d x = \frac{52}{3}

\int_{1}^{9} \frac{1}{x} d x = 4

= \frac{52}{3} + 4

Simplify

\frac{52}{3} + 4 : \frac{64}{3}

= \frac{64}{3}

What is the integral from 1 to 9 of (x+1)/(sqrt(x)) ?
The integral from 1 to 9 of (x+1)/(sqrt(x)) is 64/3