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

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

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

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

\frac{1}{2} ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 4 ∣ + C

Solution steps

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

Simplify

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

= \int \frac{x + 2}{x ( x + 4 )} d x

Expand

\frac{x + 2}{x ( x + 4 )} : \frac{x}{x ( x + 4 )} + \frac{2}{x ( 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}{x ( x + 4 )} d x + \int \frac{2}{x ( x + 4 )} d x

\int \frac{x}{x ( x + 4 )} d x = ln ∣ x + 4 ∣

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

= ln ∣ x + 4 ∣ + 2 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 4 ∣)

Simplify

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

= \frac{1}{2} ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 4 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 4 ∣ + C

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\int \frac{x}{y} d (\frac{x}{y})

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