What is the integral of (x-39)/(x^2-x-12) ?

The integral of (x-39)/(x^2-x-12) is 6ln|x+3|-5ln|x-4|+C

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integral of (x-39)/(x^2-x-12)

\int \frac{x - 39}{x ^{2} - x - 12} d x

6 ln ∣ x + 3 ∣ - 5 ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{x - 39}{x ^{2} - x - 12} d x

Take the partial fraction of

\frac{x - 39}{x ^{2} - x - 12} : \frac{6}{x + 3} - \frac{5}{x - 4}

= \int \frac{6}{x + 3} - \frac{5}{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{6}{x + 3} d x - \int \frac{5}{x - 4} d x

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

\int \frac{5}{x - 4} d x = 5 ln ∣ x - 4 ∣

= 6 ln ∣ x + 3 ∣ - 5 ln ∣ x - 4 ∣

Add a constant to the solution

= 6 ln ∣ x + 3 ∣ - 5 ln ∣ x - 4 ∣ + C

\int_{0}^{1} \frac{4 t ^{3}}{1 + t ^{4}} d t

y^{^{''}} + 9 y = 0, y (0) = - 6, y^{^{'}} (0) = - 9

\frac{d}{d x} (\frac{- 8 x}{( x ^{2} - 4 ) ^{2}})

\int_{0}^{1} \frac{1}{e ^{x}} d x

e^{x^{2} y}

What is the integral of (x-39)/(x^2-x-12) ?
The integral of (x-39)/(x^2-x-12) is 6ln|x+3|-5ln|x-4|+C