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

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

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integral of (2x+1)/(2sqrt(4x-1))

\int \frac{2 x + 1}{2 4 x - 1} d x

\frac{1}{6} (x 4 x - 1 + 2 4 x - 1) + C

Solution steps

\int \frac{2 x + 1}{2 4 x - 1} d x

Take the constant out:

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

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

Expand

\frac{2 x + 1}{4 x - 1} : \frac{2 x}{4 x - 1} + \frac{1}{4 x - 1}

= \frac{1}{2} \cdot \int \frac{2 x}{4 x - 1} + \frac{1}{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

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

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

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

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

Simplify

\frac{1}{2} (\frac{1}{6} (2 x + 1) 4 x - 1 + \frac{1}{2} 4 x - 1) : \frac{1}{6} (x 4 x - 1 + 2 4 x - 1)

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

Add a constant to the solution

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

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