What is the integral of (6x+3)/(sqrt(4x-1)) ?

The integral of (6x+3)/(sqrt(4x-1)) is xsqrt(4x-1)+2sqrt(4x-1)+C

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integral of (6x+3)/(sqrt(4x-1))

\int \frac{6 x + 3}{4 x - 1} d x

x 4 x - 1 + 2 4 x - 1 + C

Solution steps

\int \frac{6 x + 3}{4 x - 1} d x

Expand

\frac{6 x + 3}{4 x - 1} : \frac{6 x}{4 x - 1} + \frac{3}{4 x - 1}

= \int \frac{6 x}{4 x - 1} + \frac{3}{4 x - 1} d x

Apply the Sum Rule:

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

= \int \frac{6 x}{4 x - 1} d x + \int \frac{3}{4 x - 1} d x

\int \frac{6 x}{4 x - 1} d x = \frac{1}{2} (2 x + 1) 4 x - 1

\int \frac{3}{4 x - 1} d x = \frac{3}{2} 4 x - 1

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

Simplify

\frac{1}{2} (2 x + 1) 4 x - 1 + \frac{3}{2} 4 x - 1 : x 4 x - 1 + 2 4 x - 1

= x 4 x - 1 + 2 4 x - 1

Add a constant to the solution

= x 4 x - 1 + 2 4 x - 1 + C

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What is the integral of (6x+3)/(sqrt(4x-1)) ?
The integral of (6x+3)/(sqrt(4x-1)) is xsqrt(4x-1)+2sqrt(4x-1)+C