What is the integral from 0 to 24 of x/(sqrt(9+3x)) ?

The integral from 0 to 24 of x/(sqrt(9+3x)) is 40

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integral from 0 to 24 of x/(sqrt(9+3x))

\int_{0}^{24} \frac{x}{9 + 3 x} d x

40

Solution steps

\int_{0}^{24} \frac{x}{9 + 3 x} d x

Apply u-substitution

: \int_{3}^{9} \frac{2 ( u ^{2} - 9 )}{9} d u

= \int_{3}^{9} \frac{2 ( u ^{2} - 9 )}{9} d u

Take the constant out:

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

= \frac{2}{9} \cdot \int_{3}^{9} u^{2} - 9 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{2}{9} (\int_{3}^{9} u^{2} d u - \int_{3}^{9} 9 d u)

\int_{3}^{9} u^{2} d u = 234

\int_{3}^{9} 9 d u = 54

= \frac{2}{9} (234 - 54)

Simplify

\frac{2}{9} (234 - 54) : 40

= 40

x \to 0 lim (\frac{sin ( 1 )}{x})

ln (x + 3), - 1

\frac{d}{d t} ((t^{4} - 2 t)^{2} - 5)

\int_{0}^{9} 81 - x^{2} d x

x^{y} = y^{x}

What is the integral from 0 to 24 of x/(sqrt(9+3x)) ?
The integral from 0 to 24 of x/(sqrt(9+3x)) is 40