What is the integral from 9 to 16 of (sqrt(x))/(x-4) ?

The integral from 9 to 16 of (sqrt(x))/(x-4) is -4((ln(3)-ln(5))/2-1/2)

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integral from 9 to 16 of (sqrt(x))/(x-4)

\int_{9}^{16} \frac{x}{x - 4} d x

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

Decimal Notation

3.02165 \dots

Solution steps

\int_{9}^{16} \frac{x}{x - 4} d x

Apply u-substitution

: \int_{3}^{4} \frac{2 u ^{2}}{u ^{2} - 4} d u

= \int_{3}^{4} \frac{2 u ^{2}}{u ^{2} - 4} d u

Take the constant out:

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

= 2 \cdot \int_{3}^{4} \frac{u ^{2}}{u ^{2} - 4} d u

Factor

u^{2} - 4 : - (- u^{2} + 4)

= 2 \cdot \int_{3}^{4} \frac{u ^{2}}{- ( - u ^{2} + 4 )} d u

Take the constant out:

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

= 2 (- \int_{3}^{4} \frac{u ^{2}}{- u ^{2} + 4} d u)

Apply Integral Substitution

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

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

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

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

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

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

Simplify

= - 4 (\frac{ln ( 3 ) - ln ( 5 )}{2} - \frac{1}{2})

f (x) = ∣ 4 + 7 x ∣

\frac{2 x - 1}{2 x + 1}

f (x) = (x + 6)^{2}

\frac{d}{d x} (a x + ln (x))

What is the integral from 9 to 16 of (sqrt(x))/(x-4) ?
The integral from 9 to 16 of (sqrt(x))/(x-4) is -4((ln(3)-ln(5))/2-1/2)