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

The integral from 0 to 9 of (sqrt(x))/(x+9) is 6(-pi/4+1)

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

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

6 (- \frac{π}{4} + 1)

Decimal

1.28761 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{3} \frac{2 u ^{2}}{u ^{2} + 9} d u

= \int_{0}^{3} \frac{2 u ^{2}}{u ^{2} + 9} d u

Take the constant out:

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

= 2 \cdot \int_{0}^{3} \frac{u ^{2}}{u ^{2} + 9} d u

\frac{u ^{2}}{u ^{2} + 9} = - \frac{9}{u ^{2} + 9} + 1

= 2 \cdot \int_{0}^{3} - \frac{9}{u ^{2} + 9} + 1 d u

Apply Integral Substitution

= 2 \cdot \int_{0}^{1} 3 (- \frac{1}{v ^{2} + 1} + 1) d v

Take the constant out:

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

= 2 \cdot 3 \cdot \int_{0}^{1} - \frac{1}{v ^{2} + 1} + 1 d v

Apply the Sum Rule:

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

= 2 \cdot 3 (- \int_{0}^{1} \frac{1}{v ^{2} + 1} d v + \int_{0}^{1} 1 d v)

\int_{0}^{1} \frac{1}{v ^{2} + 1} d v = \frac{π}{4}

\int_{0}^{1} 1 d v = 1

= 2 \cdot 3 (- \frac{π}{4} + 1)

Simplify

= 6 (- \frac{π}{4} + 1)

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What is the integral from 0 to 9 of (sqrt(x))/(x+9) ?
The integral from 0 to 9 of (sqrt(x))/(x+9) is 6(-pi/4+1)