What is the integral of (sqrt(x+9))/x ?

The integral of (sqrt(x+9))/x is -6(1/2 ln|(sqrt(x+9))/3+1|-1/2 ln|(sqrt(x+9))/3-1|-1/3 sqrt(x+9))+C

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integral of (sqrt(x+9))/x

\int \frac{x + 9}{x} d x

- 6 (\frac{1}{2} ln \frac{x + 9}{3} + 1 - \frac{1}{2} ln \frac{x + 9}{3} - 1 - \frac{1}{3} x + 9) + C

Solution steps

\int \frac{x + 9}{x} d x

Apply u-substitution

: \int \frac{2 u ^{2}}{u ^{2} - 9} d u

= \int \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 \frac{u ^{2}}{u ^{2} - 9} d u

Factor

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

= 2 \cdot \int \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 \frac{u ^{2}}{- u ^{2} + 9} d u)

Apply Integral Substitution

= 2 (- \int \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{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}{- 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}{- 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 (- 3 (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2} - v))

Substitute back

= 2 - 3 \frac{ln \frac{x + 9}{3} + 1}{2} - \frac{ln \frac{x + 9}{3} - 1}{2} - \frac{x + 9}{3}

Simplify

2 - 3 \frac{ln \frac{x + 9}{3} + 1}{2} - \frac{ln \frac{x + 9}{3} - 1}{2} - \frac{x + 9}{3} : - 6 (\frac{1}{2} ln \frac{x + 9}{3} + 1 - \frac{1}{2} ln \frac{x + 9}{3} - 1 - \frac{1}{3} x + 9)

= - 6 (\frac{1}{2} ln \frac{x + 9}{3} + 1 - \frac{1}{2} ln \frac{x + 9}{3} - 1 - \frac{1}{3} x + 9)

Add a constant to the solution

= - 6 (\frac{1}{2} ln \frac{x + 9}{3} + 1 - \frac{1}{2} ln \frac{x + 9}{3} - 1 - \frac{1}{3} x + 9) + C

(y^{2} + 1) d x = (y sec^{2} (x)) d y

\int \frac{x ^{3}}{9 x ^{2} + 49} d x

(\frac{2 t + 1}{2 t})^{^{'}}

\frac{\partial}{\partial r} (arctan (rs))

\frac{d ^{5}}{d x ^{5}} (ln (x + 1))

What is the integral of (sqrt(x+9))/x ?
The integral of (sqrt(x+9))/x is -6(1/2 ln|(sqrt(x+9))/3+1|-1/2 ln|(sqrt(x+9))/3-1|-1/3 sqrt(x+9))+C