What is the integral from 4 to 9 of (sqrt(x))/(x-36) ?

The integral from 4 to 9 of (sqrt(x))/(x-36) is -12(1/2 ln(3/2)-1/6)

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integral from 4 to 9 of (sqrt(x))/(x-36)

\int_{4}^{9} \frac{x}{x - 36} d x

- 12 (\frac{1}{2} ln (\frac{3}{2}) - \frac{1}{6})

Decimal

- 0.43279 \dots

Solution steps

\int_{4}^{9} \frac{x}{x - 36} d x

Apply u-substitution

: \int_{2}^{3} \frac{2 u ^{2}}{u ^{2} - 36} d u

= \int_{2}^{3} \frac{2 u ^{2}}{u ^{2} - 36} d u

Take the constant out:

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

= 2 \cdot \int_{2}^{3} \frac{u ^{2}}{u ^{2} - 36} d u

Factor

u^{2} - 36 : - (- u^{2} + 36)

= 2 \cdot \int_{2}^{3} \frac{u ^{2}}{- ( - u ^{2} + 36 )} d u

Take the constant out:

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

= 2 (- \int_{2}^{3} \frac{u ^{2}}{- u ^{2} + 36} d u)

Apply Integral Substitution

= 2 (- \int_{\frac{1}{3}}^{\frac{1}{2}} \frac{6 v ^{2}}{- v ^{2} + 1} d v)

Take the constant out:

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

= 2 (- 6 \cdot \int_{\frac{1}{3}}^{\frac{1}{2}} \frac{v ^{2}}{- v ^{2} + 1} d v)

\frac{v ^{2}}{- v ^{2} + 1} = \frac{1}{- v ^{2} + 1} - 1

= 2 (- 6 \cdot \int_{\frac{1}{3}}^{\frac{1}{2}} \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 (- 6 (\int_{\frac{1}{3}}^{\frac{1}{2}} \frac{1}{- v ^{2} + 1} d v - \int_{\frac{1}{3}}^{\frac{1}{2}} 1 d v))

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

\int_{\frac{1}{3}}^{\frac{1}{2}} 1 d v = \frac{1}{6}

= 2 (- 6 (\frac{ln ( \frac{3}{2} )}{2} - \frac{1}{6}))

Simplify

2 (- 6 (\frac{ln ( \frac{3}{2} )}{2} - \frac{1}{6})) : - 12 (\frac{1}{2} ln (\frac{3}{2}) - \frac{1}{6})

= - 12 (\frac{1}{2} ln (\frac{3}{2}) - \frac{1}{6})

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What is the integral from 4 to 9 of (sqrt(x))/(x-36) ?
The integral from 4 to 9 of (sqrt(x))/(x-36) is -12(1/2 ln(3/2)-1/6)