What is the integral from 4 to 9 of ((sqrt(x+3))^2)/(2sqrt(x)) ?

The integral from 4 to 9 of ((sqrt(x+3))^2)/(2sqrt(x)) is 28/3

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integral from 4 to 9 of ((sqrt(x+3))^2)/(2sqrt(x))

\int_{4}^{9} \frac{( x + 3 ) ^{2}}{2 x} d x

\frac{28}{3}

Solution steps

\int_{4}^{9} \frac{( x + 3 ) ^{2}}{2 x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{4}^{9} \frac{( x + 3 ) ^{2}}{x} d x

(x + 3)^{2} = x + 3

= \frac{1}{2} \cdot \int_{4}^{9} \frac{x + 3}{x} d x

Expand

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

= \frac{1}{2} \cdot \int_{4}^{9} x + \frac{3}{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

= \frac{1}{2} (\int_{4}^{9} x d x + \int_{4}^{9} \frac{3}{x} d x)

\int_{4}^{9} x d x = \frac{38}{3}

\int_{4}^{9} \frac{3}{x} d x = 6

= \frac{1}{2} (\frac{38}{3} + 6)

Simplify

\frac{1}{2} (\frac{38}{3} + 6) : \frac{28}{3}

= \frac{28}{3}

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What is the integral from 4 to 9 of ((sqrt(x+3))^2)/(2sqrt(x)) ?
The integral from 4 to 9 of ((sqrt(x+3))^2)/(2sqrt(x)) is 28/3