What is the integral of 1/((x+3)(x-3)) ?

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

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integral of 1/((x+3)(x-3))

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

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

Solution steps

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

Take the partial fraction of

\frac{1}{( x + 3 ) ( x - 3 )} : - \frac{1}{6 ( x + 3 )} + \frac{1}{6 ( x - 3 )}

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

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

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

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

Add a constant to the solution

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

\frac{d}{d x} (x - 3 sin (x))

\int \frac{e ^{u}}{( 4 - e ^{u} ) ^{2}} d u

\frac{d y}{d x} = y^{2} e^{- x}

\int x \cdot x^{20} d x

t^{2} e^{- 2 t}

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