What is the integral of (sqrt(3x)+4)/(sqrt(3x)-2) ?

The integral of (sqrt(3x)+4)/(sqrt(3x)-2) is 1/3 (3x+12sqrt(3)sqrt(x)+24ln|sqrt(3)sqrt(x)-2|-28)+C

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integral of (sqrt(3x)+4)/(sqrt(3x)-2)

\int \frac{3 x + 4}{3 x - 2} d x

\frac{1}{3} (3 x + 123 x + 24 ln 3 x - 2 - 28) + C

Solution steps

\int \frac{3 x + 4}{3 x - 2} d x

Apply u-substitution

: \int \frac{u + 4}{3 ( u - 2 )} d u

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

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{u + 4}{u - 2} d u

Expand

\frac{u + 4}{u - 2} : \frac{u}{u - 2} + \frac{4}{u - 2}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int \frac{u}{u - 2} d u + \int \frac{4}{u - 2} d u)

\int \frac{u}{u - 2} d u = u + 4 u + 8 ln u - 2 - 12

\int \frac{4}{u - 2} d u = 8 (u - 2 + 2 ln u - 2)

= \frac{1}{3} (u + 4 u + 8 ln u - 2 - 12 + 8 (u - 2 + 2 ln u - 2))

Substitute back

u = 3 x

= \frac{1}{3} (3 x + 4 3 x + 8 ln 3 x - 2 - 12 + 8 (3 x - 2 + 2 ln 3 x - 2))

Simplify

\frac{1}{3} (3 x + 4 3 x + 8 ln 3 x - 2 - 12 + 8 (3 x - 2 + 2 ln 3 x - 2)) : \frac{1}{3} (3 x + 123 x + 24 ln 3 x - 2 - 28)

= \frac{1}{3} (3 x + 123 x + 24 ln 3 x - 2 - 28)

Add a constant to the solution

= \frac{1}{3} (3 x + 123 x + 24 ln 3 x - 2 - 28) + C

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What is the integral of (sqrt(3x)+4)/(sqrt(3x)-2) ?
The integral of (sqrt(3x)+4)/(sqrt(3x)-2) is 1/3 (3x+12sqrt(3)sqrt(x)+24ln|sqrt(3)sqrt(x)-2|-28)+C