What is the integral of 1/(x^2-3x-4) ?

The integral of 1/(x^2-3x-4) is -1/5 ln|x+1|+1/5 ln|x-4|+C

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integral of 1/(x^2-3x-4)

\int \frac{1}{x ^{2} - 3 x - 4} d x

- \frac{1}{5} ln ∣ x + 1 ∣ + \frac{1}{5} ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{1}{x ^{2} - 3 x - 4} d x

Take the partial fraction of

\frac{1}{x ^{2} - 3 x - 4} : - \frac{1}{5 ( x + 1 )} + \frac{1}{5 ( x - 4 )}

= \int - \frac{1}{5 ( x + 1 )} + \frac{1}{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{1}{5 ( x + 1 )} d x + \int \frac{1}{5 ( x - 4 )} d x

\int \frac{1}{5 ( x + 1 )} d x = \frac{1}{5} ln ∣ x + 1 ∣

\int \frac{1}{5 ( x - 4 )} d x = \frac{1}{5} ln ∣ x - 4 ∣

= - \frac{1}{5} ln ∣ x + 1 ∣ + \frac{1}{5} ln ∣ x - 4 ∣

Add a constant to the solution

= - \frac{1}{5} ln ∣ x + 1 ∣ + \frac{1}{5} ln ∣ x - 4 ∣ + C

\int (x + 1)^{\frac{3}{2}} d x

\frac{d}{d x} (e^{t a n^{2} (x)})

\frac{\partial}{\partial x} (x + 9 y + 8 z)

\frac{d}{d x} (\frac{7}{3 x ^{7} - 12 x ^{6} + 2 x ^{2} - 9})

y = \frac{1}{x + 2}, at x = 7

What is the integral of 1/(x^2-3x-4) ?
The integral of 1/(x^2-3x-4) is -1/5 ln|x+1|+1/5 ln|x-4|+C