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

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

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

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

\frac{3}{4} ln ∣ x ∣ - \frac{3}{4} ln ∣ x + 4 ∣ + \frac{1}{x} + C

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

= \frac{3}{4} ln ∣ x ∣ - \frac{3}{4} ln ∣ x + 4 ∣ + \frac{1}{x} + C

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