What is the integral of (x^2+x+8)/(x^2+3x-4) ?

The integral of (x^2+x+8)/(x^2+3x-4) is x+2ln|x-1|-4ln|x+4|+C

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integral of (x^2+x+8)/(x^2+3x-4)

\int \frac{x ^{2} + x + 8}{x ^{2} + 3 x - 4} d x

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

Solution steps

\int \frac{x ^{2} + x + 8}{x ^{2} + 3 x - 4} d x

Expand

\frac{x ^{2} + x + 8}{x ^{2} + 3 x - 4} : \frac{x ^{2}}{x ^{2} + 3 x - 4} + \frac{x}{x ^{2} + 3 x - 4} + \frac{8}{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{x ^{2}}{x ^{2} + 3 x - 4} d x + \int \frac{x}{x ^{2} + 3 x - 4} d x + \int \frac{8}{x ^{2} + 3 x - 4} d x

\int \frac{x ^{2}}{x ^{2} + 3 x - 4} d x = x + \frac{1}{5} ln ∣ x - 1 ∣ - \frac{16}{5} ln ∣ x + 4 ∣

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

\int \frac{8}{x ^{2} + 3 x - 4} d x = 8 (\frac{1}{5} ln ∣ x - 1 ∣ - \frac{1}{5} ln ∣ x + 4 ∣)

= x + \frac{1}{5} ln ∣ x - 1 ∣ - \frac{16}{5} ln ∣ x + 4 ∣ + \frac{1}{5} ln ∣ x - 1 ∣ + \frac{4}{5} ln ∣ x + 4 ∣ + 8 (\frac{1}{5} ln ∣ x - 1 ∣ - \frac{1}{5} ln ∣ x + 4 ∣)

Simplify

x + \frac{1}{5} ln ∣ x - 1 ∣ - \frac{16}{5} ln ∣ x + 4 ∣ + \frac{1}{5} ln ∣ x - 1 ∣ + \frac{4}{5} ln ∣ x + 4 ∣ + 8 (\frac{1}{5} ln ∣ x - 1 ∣ - \frac{1}{5} ln ∣ x + 4 ∣) : x + 2 ln ∣ x - 1 ∣ - 4 ln ∣ x + 4 ∣

= x + 2 ln ∣ x - 1 ∣ - 4 ln ∣ x + 4 ∣

Add a constant to the solution

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

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What is the integral of (x^2+x+8)/(x^2+3x-4) ?
The integral of (x^2+x+8)/(x^2+3x-4) is x+2ln|x-1|-4ln|x+4|+C