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

The integral from 0 to 3 of sqrt(x/(9-x)) is 3(3arctan(sqrt(1/2))-sqrt(2))

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

\int_{0}^{3} \frac{x}{9 - x} d x

3 (3 arctan (\frac{1}{2}) - 2)

Decimal Notation

1.29667 \dots

Solution steps

\int_{0}^{3} \frac{x}{9 - x} d x

Apply u-substitution

: \int_{9}^{6} - \frac{- u + 9}{u} d u

= \int_{9}^{6} - \frac{- u + 9}{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{6}^{9} - \frac{- u + 9}{u} d u

Take the constant out:

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

= - (- \int_{6}^{9} \frac{- u + 9}{u} d u)

Expand

\frac{- u + 9}{u} : - 1 + \frac{9}{u}

= - (- \int_{6}^{9} - 1 + \frac{9}{u} d u)

Apply u-substitution

: \int_{\frac{3}{2}}^{1} - \frac{9 v - 1}{v ^{2}} d v

= - (- \int_{\frac{3}{2}}^{1} - \frac{9 v - 1}{v ^{2}} d v)

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - (- (- \int_{1}^{\frac{3}{2}} - \frac{9 v - 1}{v ^{2}} d v))

Take the constant out:

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

= - (- (- (- 9 \cdot \int_{1}^{\frac{3}{2}} \frac{v - 1}{v ^{2}} d v)))

Apply Integration By Parts

: [\frac{- v - 1 - v \cdot \int - \frac{1}{2 v v - 1} d v}{v}]_{1}^{\frac{3}{2}}

= - - - - 9 [\frac{- v - 1 - v \cdot \int - \frac{1}{2 v v - 1} d v}{v}]_{1}^{\frac{3}{2}}

\int - \frac{1}{2 v v - 1} d v = - arctan (v - 1)

= - - - - 9 [\frac{- v - 1 - v ( - arctan ( v - 1 ) )}{v}]_{1}^{\frac{3}{2}}

Simplify

= 9 [\frac{- v - 1 + v arctan ( v - 1 )}{v}]_{1}^{\frac{3}{2}}

Compute the boundaries:

\frac{3 arctan ( \frac{1}{2} ) - 2}{3}

= 9 \cdot \frac{3 arctan ( \frac{1}{2} ) - 2}{3}

Simplify

= 3 (3 arctan (\frac{1}{2}) - 2)

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What is the integral from 0 to 3 of sqrt(x/(9-x)) ?
The integral from 0 to 3 of sqrt(x/(9-x)) is 3(3arctan(sqrt(1/2))-sqrt(2))