What is the integral of (4x^2-7x-19)/(x^2-2x-3) ?

The integral of (4x^2-7x-19)/(x^2-2x-3) is 4x+2ln|x+1|-ln|x-3|-4+C

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integral of (4x^2-7x-19)/(x^2-2x-3)

\int \frac{4 x ^{2} - 7 x - 19}{x ^{2} - 2 x - 3} d x

4 x + 2 ln ∣ x + 1 ∣ - ln ∣ x - 3 ∣ - 4 + C

Solution steps

\int \frac{4 x ^{2} - 7 x - 19}{x ^{2} - 2 x - 3} d x

Expand

\frac{4 x ^{2} - 7 x - 19}{x ^{2} - 2 x - 3} : \frac{4 x ^{2}}{x ^{2} - 2 x - 3} - \frac{7 x}{x ^{2} - 2 x - 3} - \frac{19}{x ^{2} - 2 x - 3}

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{4 x ^{2}}{x ^{2} - 2 x - 3} d x - \int \frac{7 x}{x ^{2} - 2 x - 3} d x - \int \frac{19}{x ^{2} - 2 x - 3} d x

\int \frac{4 x ^{2}}{x ^{2} - 2 x - 3} d x = 4 x - 4 - ln ∣ x + 1 ∣ + 9 ln ∣ x - 3 ∣

\int \frac{7 x}{x ^{2} - 2 x - 3} d x = 7 (\frac{1}{4} ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ x - 3 ∣)

\int \frac{19}{x ^{2} - 2 x - 3} d x = 19 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣)

= 4 x - 4 - ln ∣ x + 1 ∣ + 9 ln ∣ x - 3 ∣ - 7 (\frac{1}{4} ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ x - 3 ∣) - 19 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣)

Simplify

4 x - 4 - ln ∣ x + 1 ∣ + 9 ln ∣ x - 3 ∣ - 7 (\frac{1}{4} ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ x - 3 ∣) - 19 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣) : 4 x + 2 ln ∣ x + 1 ∣ - ln ∣ x - 3 ∣ - 4

= 4 x + 2 ln ∣ x + 1 ∣ - ln ∣ x - 3 ∣ - 4

Add a constant to the solution

= 4 x + 2 ln ∣ x + 1 ∣ - ln ∣ x - 3 ∣ - 4 + C

- 3 a e^{- 3 x} + b e^{x} + c (e^{x} + e^{x} x)

\frac{1}{2} (e^{(1 - x)})

y = - \frac{12}{7} x = + b

\frac{d}{d x} y = tan^{2} (x)

\int e^{- x} sin (x) d x

What is the integral of (4x^2-7x-19)/(x^2-2x-3) ?
The integral of (4x^2-7x-19)/(x^2-2x-3) is 4x+2ln|x+1|-ln|x-3|-4+C