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

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

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

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

\frac{1}{2} ln ∣ x + 3 ∣ + \frac{3}{2} ln ∣ x - 3 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{2 x + 3}{x ^{2} - 9} : \frac{1}{2 ( x + 3 )} + \frac{3}{2 ( x - 3 )}

= \int \frac{1}{2 ( x + 3 )} + \frac{3}{2 ( x - 3 )} 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{1}{2 ( x + 3 )} d x + \int \frac{3}{2 ( x - 3 )} d x

\int \frac{1}{2 ( x + 3 )} d x = \frac{1}{2} ln ∣ x + 3 ∣

\int \frac{3}{2 ( x - 3 )} d x = \frac{3}{2} ln ∣ x - 3 ∣

= \frac{1}{2} ln ∣ x + 3 ∣ + \frac{3}{2} ln ∣ x - 3 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ x + 3 ∣ + \frac{3}{2} ln ∣ x - 3 ∣ + C

\frac{d}{d x} (x 25 - x^{2})

\int \frac{3 x - 4}{( x - 1 ) ( x + 2 )} d x

\frac{1}{s ( s + 1 ) ^{3}}

x^{^{''}} + 2 x^{^{'}} + 5 x = 0, x (0) = 1, x^{^{'}} (0) = - 1

y = 8 x + 3 x^{\frac{1}{6}}

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