What is the integral of (sqrt(x))/(x-16) ?

The integral of (sqrt(x))/(x-16) is -8(1/2 ln|(sqrt(x))/4+1|-1/2 ln|(sqrt(x))/4-1|-1/4 sqrt(x))+C

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integral of (sqrt(x))/(x-16)

\int \frac{x}{x - 16} d x

- 8 (\frac{1}{2} ln \frac{x}{4} + 1 - \frac{1}{2} ln \frac{x}{4} - 1 - \frac{1}{4} x) + C

Solution steps

\int \frac{x}{x - 16} d x

Apply u-substitution

: \int \frac{2 u ^{2}}{u ^{2} - 16} d u

= \int \frac{2 u ^{2}}{u ^{2} - 16} d u

Take the constant out:

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

= 2 \cdot \int \frac{u ^{2}}{u ^{2} - 16} d u

Factor

u^{2} - 16 : - (- u^{2} + 16)

= 2 \cdot \int \frac{u ^{2}}{- ( - u ^{2} + 16 )} d u

Take the constant out:

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

= 2 (- \int \frac{u ^{2}}{- u ^{2} + 16} d u)

Apply Integral Substitution

= 2 (- \int \frac{4 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 (- 4 \cdot \int \frac{v ^{2}}{- v ^{2} + 1} d v)

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

= 2 (- 4 \cdot \int \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 (- 4 (\int \frac{1}{- v ^{2} + 1} d v - \int 1 d v))

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

\int 1 d v = v

= 2 (- 4 (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2} - v))

Substitute back

= 2 - 4 \frac{ln \frac{x}{4} + 1}{2} - \frac{ln \frac{x}{4} - 1}{2} - \frac{x}{4}

Simplify

2 - 4 \frac{ln \frac{x}{4} + 1}{2} - \frac{ln \frac{x}{4} - 1}{2} - \frac{x}{4} : - 8 (\frac{1}{2} ln \frac{x}{4} + 1 - \frac{1}{2} ln \frac{x}{4} - 1 - \frac{1}{4} x)

= - 8 (\frac{1}{2} ln \frac{x}{4} + 1 - \frac{1}{2} ln \frac{x}{4} - 1 - \frac{1}{4} x)

Add a constant to the solution

= - 8 (\frac{1}{2} ln \frac{x}{4} + 1 - \frac{1}{2} ln \frac{x}{4} - 1 - \frac{1}{4} x) + C

n = 0 \sum \infty n^{2} (x)^{n - 1}

z \to - 1 lim (\frac{4 z + 3}{2 z + 1})

5 t^{4}

x^{7} y^{^{'}} = - 8 x^{6} y + cos (2 x)

\int \frac{- 2 a x}{x ^{2} + y ^{2}} d y

What is the integral of (sqrt(x))/(x-16) ?
The integral of (sqrt(x))/(x-16) is -8(1/2 ln|(sqrt(x))/4+1|-1/2 ln|(sqrt(x))/4-1|-1/4 sqrt(x))+C