What is the integral of (x-7)/(x^2-x-12) ?

The integral of (x-7)/(x^2-x-12) is 10/7 ln|x+3|-3/7 ln|x-4|+C

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integral of (x-7)/(x^2-x-12)

\int \frac{x - 7}{x ^{2} - x - 12} d x

\frac{10}{7} ln ∣ x + 3 ∣ - \frac{3}{7} ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{x - 7}{x ^{2} - x - 12} d x

Take the partial fraction of

\frac{x - 7}{x ^{2} - x - 12} : \frac{10}{7 ( x + 3 )} - \frac{3}{7 ( x - 4 )}

= \int \frac{10}{7 ( x + 3 )} - \frac{3}{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{10}{7 ( x + 3 )} d x - \int \frac{3}{7 ( x - 4 )} d x

\int \frac{10}{7 ( x + 3 )} d x = \frac{10}{7} ln ∣ x + 3 ∣

\int \frac{3}{7 ( x - 4 )} d x = \frac{3}{7} ln ∣ x - 4 ∣

= \frac{10}{7} ln ∣ x + 3 ∣ - \frac{3}{7} ln ∣ x - 4 ∣

Add a constant to the solution

= \frac{10}{7} ln ∣ x + 3 ∣ - \frac{3}{7} ln ∣ x - 4 ∣ + C

\frac{\partial}{\partial x} (8 x e^{5 x y})

n = 0 \sum \infty \frac{n !}{1 1 ^{n}}

y = 2 e^{2 e^{x} + x}

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

f (x) = 3 x^{4} + 2 x - 1, (2, 51)

What is the integral of (x-7)/(x^2-x-12) ?
The integral of (x-7)/(x^2-x-12) is 10/7 ln|x+3|-3/7 ln|x-4|+C