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

The integral of (2x+3)/((x-3)(x+3)) is ln(|x-3||x+3|)+3(1/6 ln|x-3|-1/6 ln|x+3|)+C

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

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

ln (∣ x - 3 ∣ ∣ x + 3 ∣) + 3 (\frac{1}{6} ln ∣ x - 3 ∣ - \frac{1}{6} ln ∣ x + 3 ∣) + C

Solution steps

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

Expand

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

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

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

\int \frac{3}{( x - 3 ) ( x + 3 )} d x = 3 (\frac{1}{6} ln ∣ x - 3 ∣ - \frac{1}{6} ln ∣ x + 3 ∣)

= ln (∣ x - 3 ∣ ∣ x + 3 ∣) + 3 (\frac{1}{6} ln ∣ x - 3 ∣ - \frac{1}{6} ln ∣ x + 3 ∣)

Add a constant to the solution

= ln (∣ x - 3 ∣ ∣ x + 3 ∣) + 3 (\frac{1}{6} ln ∣ x - 3 ∣ - \frac{1}{6} ln ∣ x + 3 ∣) + C

\frac{\partial}{\partial x} (tan (y) e^{x})

x \to 1 lim (x^{2} - 2 x + 1)

n = 0 \sum \infty 6^{n + 1}

n = 0 \sum \infty (- 1)^{n} e^{- n}

\int e^{9 x} cos (x) d x

What is the integral of (2x+3)/((x-3)(x+3)) ?
The integral of (2x+3)/((x-3)(x+3)) is ln(|x-3||x+3|)+3(1/6 ln|x-3|-1/6 ln|x+3|)+C