What is the integral of (sqrt(x))/(x-4) ?

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

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integral of (sqrt(x))/(x-4)

\int \frac{x}{x - 4} d x

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

Solution steps

\int \frac{x}{x - 4} d x

Apply u-substitution

: \int \frac{2 u ^{2}}{u ^{2} - 4} d u

= \int \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 \frac{u ^{2}}{u ^{2} - 4} d u

Factor

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

= 2 \cdot \int \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 \frac{u ^{2}}{- u ^{2} + 4} d u)

Apply Integral Substitution

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

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

f (x) = x, 5, at x = 25

x \to e lim (3 ln (x) - 1)

\int_{1}^{4} \frac{1}{t} d t

\frac{d}{d x} (\frac{2 x ^{3} - 16}{x - 2})

y = \frac{5 x}{x + 3}, (2, 2)

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