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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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