What is the integral of (2x-1)/(sqrt(9x^2-4)) ?

The integral of (2x-1)/(sqrt(9x^2-4)) is 2/9 sqrt(9x^2-4)-1/3 ln((|sqrt(9x^2-4)+3x|)/2)+C

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integral of (2x-1)/(sqrt(9x^2-4))

\int \frac{2 x - 1}{9 x ^{2} - 4} d x

\frac{2}{9} 9 x^{2} - 4 - \frac{1}{3} ln (\frac{9 x ^{2} - 4 + 3 x}{2}) + C

Solution steps

\int \frac{2 x - 1}{9 x ^{2} - 4} d x

Expand

\frac{2 x - 1}{9 x ^{2} - 4} : \frac{2 x}{9 x ^{2} - 4} - \frac{1}{9 x ^{2} - 4}

= \int \frac{2 x}{9 x ^{2} - 4} - \frac{1}{9 x ^{2} - 4} 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{2 x}{9 x ^{2} - 4} d x - \int \frac{1}{9 x ^{2} - 4} d x

\int \frac{2 x}{9 x ^{2} - 4} d x = \frac{2}{9} 9 x^{2} - 4

\int \frac{1}{9 x ^{2} - 4} d x = \frac{1}{3} ln (\frac{9 x ^{2} - 4 + 3 x}{2})

= \frac{2}{9} 9 x^{2} - 4 - \frac{1}{3} ln (\frac{9 x ^{2} - 4 + 3 x}{2})

Add a constant to the solution

= \frac{2}{9} 9 x^{2} - 4 - \frac{1}{3} ln (\frac{9 x ^{2} - 4 + 3 x}{2}) + C

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What is the integral of (2x-1)/(sqrt(9x^2-4)) ?
The integral of (2x-1)/(sqrt(9x^2-4)) is 2/9 sqrt(9x^2-4)-1/3 ln((|sqrt(9x^2-4)+3x|)/2)+C