What is the integral from 1 to 4 of (sqrt(x))/(x-9) ?

The integral from 1 to 4 of (sqrt(x))/(x-9) is -6(1/2 ln(5/2)-1/3)

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integral from 1 to 4 of (sqrt(x))/(x-9)

\int_{1}^{4} \frac{x}{x - 9} d x

- 6 (\frac{1}{2} ln (\frac{5}{2}) - \frac{1}{3})

Decimal Notation

- 0.74887 \dots

Solution steps

\int_{1}^{4} \frac{x}{x - 9} d x

Apply u-substitution

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

= \int_{1}^{2} \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_{1}^{2} \frac{u ^{2}}{u ^{2} - 9} d u

Factor

u^{2} - 9 : - (- u^{2} + 9)

= 2 \cdot \int_{1}^{2} \frac{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 (- \int_{1}^{2} \frac{u ^{2}}{- u ^{2} + 9} d u)

Apply Integral Substitution

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

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

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

\int_{\frac{1}{3}}^{\frac{2}{3}} \frac{1}{- v ^{2} + 1} d v = \frac{ln ( \frac{5}{2} )}{2}

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

= 2 (- 3 (\frac{ln ( \frac{5}{2} )}{2} - \frac{1}{3}))

Simplify

2 (- 3 (\frac{ln ( \frac{5}{2} )}{2} - \frac{1}{3})) : - 6 (\frac{1}{2} ln (\frac{5}{2}) - \frac{1}{3})

= - 6 (\frac{1}{2} ln (\frac{5}{2}) - \frac{1}{3})

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\frac{d}{d x} (\frac{x}{2 + x})

\frac{\partial}{\partial x} (\frac{3}{x ^{2} + z ^{2}})

\frac{\partial}{\partial x} (x^{3} y - y^{3})

What is the integral from 1 to 4 of (sqrt(x))/(x-9) ?
The integral from 1 to 4 of (sqrt(x))/(x-9) is -6(1/2 ln(5/2)-1/3)