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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Add a constant to the solution

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

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