What is the integral of (sqrt(x))/(x-36) ?

The integral of (sqrt(x))/(x-36) is -12(1/2 ln|(sqrt(x))/6+1|-1/2 ln|(sqrt(x))/6-1|-1/6 sqrt(x))+C

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integral of (sqrt(x))/(x-36)

\int \frac{x}{x - 36} d x

- 12 (\frac{1}{2} ln \frac{x}{6} + 1 - \frac{1}{2} ln \frac{x}{6} - 1 - \frac{1}{6} x) + C

Solution steps

\int \frac{x}{x - 36} d x

Apply u-substitution

: \int \frac{2 u ^{2}}{u ^{2} - 36} d u

= \int \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 \frac{u ^{2}}{u ^{2} - 36} d u

Factor

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

= 2 \cdot \int \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 \frac{u ^{2}}{- u ^{2} + 36} d u)

Apply Integral Substitution

= 2 (- \int \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{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}{- 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}{- 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 (- 6 (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2} - v))

Substitute back

= 2 - 6 \frac{ln \frac{x}{6} + 1}{2} - \frac{ln \frac{x}{6} - 1}{2} - \frac{x}{6}

Simplify

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

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

Add a constant to the solution

= - 12 (\frac{1}{2} ln \frac{x}{6} + 1 - \frac{1}{2} ln \frac{x}{6} - 1 - \frac{1}{6} x) + C

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What is the integral of (sqrt(x))/(x-36) ?
The integral of (sqrt(x))/(x-36) is -12(1/2 ln|(sqrt(x))/6+1|-1/2 ln|(sqrt(x))/6-1|-1/6 sqrt(x))+C