What is the integral of (2x+3)/(4x^2-9) ?

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

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integral of (2x+3)/(4x^2-9)

\int \frac{2 x + 3}{4 x ^{2} - 9} d x

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

Solution steps

\int \frac{2 x + 3}{4 x ^{2} - 9} d x

Expand

\frac{2 x + 3}{4 x ^{2} - 9} : \frac{2 x}{4 x ^{2} - 9} + \frac{3}{4 x ^{2} - 9}

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}{4 x ^{2} - 9} d x + \int \frac{3}{4 x ^{2} - 9} d x

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

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

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

Add a constant to the solution

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

x \to 0 lim (\frac{x + 2}{x ^{2}})

x \to - 4 lim (x^{2} + 3 x - 7)

\frac{\partial}{\partial y} (x z cos (x + yz))

\frac{d}{d x} (sin^{2} (ln (x)))

\frac{3 x}{9 x ^{2} + 4 x - 9}

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