What is the integral of 1/((x-3)(x+4)) ?

The integral of 1/((x-3)(x+4)) is 1/7 ln|x-3|-1/7 ln|x+4|+C

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integral of 1/((x-3)(x+4))

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

\frac{1}{7} ln ∣ x - 3 ∣ - \frac{1}{7} ln ∣ x + 4 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{( x - 3 ) ( x + 4 )} : \frac{1}{7 ( x - 3 )} - \frac{1}{7 ( x + 4 )}

= \int \frac{1}{7 ( x - 3 )} - \frac{1}{7 ( x + 4 )} 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}{7 ( x - 3 )} d x - \int \frac{1}{7 ( x + 4 )} d x

\int \frac{1}{7 ( x - 3 )} d x = \frac{1}{7} ln ∣ x - 3 ∣

\int \frac{1}{7 ( x + 4 )} d x = \frac{1}{7} ln ∣ x + 4 ∣

= \frac{1}{7} ln ∣ x - 3 ∣ - \frac{1}{7} ln ∣ x + 4 ∣

Add a constant to the solution

= \frac{1}{7} ln ∣ x - 3 ∣ - \frac{1}{7} ln ∣ x + 4 ∣ + C

x \to 0 lim (x^{0})

f (x) = \frac{1}{x - 3}

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What is the integral of 1/((x-3)(x+4)) ?
The integral of 1/((x-3)(x+4)) is 1/7 ln|x-3|-1/7 ln|x+4|+C