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

The integral of (3x-1)/(x^2-3x-4) is 4/5 ln|x+1|+11/5 ln|x-4|+C

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

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

\frac{4}{5} ln ∣ x + 1 ∣ + \frac{11}{5} ln ∣ x - 4 ∣ + C

Solution steps

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

Expand

\frac{3 x - 1}{x ^{2} - 3 x - 4} : \frac{3 x}{x ^{2} - 3 x - 4} - \frac{1}{x ^{2} - 3 x - 4}

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

\int \frac{3 x}{x ^{2} - 3 x - 4} d x = 3 (\frac{1}{5} ln ∣ x + 1 ∣ + \frac{4}{5} ln ∣ x - 4 ∣)

\int \frac{1}{x ^{2} - 3 x - 4} d x = - \frac{1}{5} ln ∣ x + 1 ∣ + \frac{1}{5} ln ∣ x - 4 ∣

= 3 (\frac{1}{5} ln ∣ x + 1 ∣ + \frac{4}{5} ln ∣ x - 4 ∣) - (- \frac{1}{5} ln ∣ x + 1 ∣ + \frac{1}{5} ln ∣ x - 4 ∣)

Simplify

3 (\frac{1}{5} ln ∣ x + 1 ∣ + \frac{4}{5} ln ∣ x - 4 ∣) - (- \frac{1}{5} ln ∣ x + 1 ∣ + \frac{1}{5} ln ∣ x - 4 ∣) : \frac{4}{5} ln ∣ x + 1 ∣ + \frac{11}{5} ln ∣ x - 4 ∣

= \frac{4}{5} ln ∣ x + 1 ∣ + \frac{11}{5} ln ∣ x - 4 ∣

Add a constant to the solution

= \frac{4}{5} ln ∣ x + 1 ∣ + \frac{11}{5} ln ∣ x - 4 ∣ + C

\int e^{2 t + e^{t}} d t

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

\int - \frac{5}{3 x} d x

d er i v a t i v e \frac{( 2 x ^{4} + 5 x ^{3} - 2 x - 8 )}{( x + 3 )}

\frac{\partial}{\partial x} (\frac{4 x - 3 t}{s + 2 x})

What is the integral of (3x-1)/(x^2-3x-4) ?
The integral of (3x-1)/(x^2-3x-4) is 4/5 ln|x+1|+11/5 ln|x-4|+C