What is the integral of (6x^2-5x+4)/(x^3+4x) ?

The integral of (6x^2-5x+4)/(x^3+4x) is -5/2 arctan(x/2)+5/2 ln|x^2+4|+ln|x|+C

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integral of (6x^2-5x+4)/(x^3+4x)

\int \frac{6 x ^{2} - 5 x + 4}{x ^{3} + 4 x} d x

- \frac{5}{2} arctan (\frac{x}{2}) + \frac{5}{2} ln x^{2} + 4 + ln ∣ x ∣ + C

Solution steps

\int \frac{6 x ^{2} - 5 x + 4}{x ^{3} + 4 x} d x

Expand

\frac{6 x ^{2} - 5 x + 4}{x ^{3} + 4 x} : \frac{6 x ^{2}}{x ^{3} + 4 x} - \frac{5 x}{x ^{3} + 4 x} + \frac{4}{x ^{3} + 4 x}

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{6 x ^{2}}{x ^{3} + 4 x} d x - \int \frac{5 x}{x ^{3} + 4 x} d x + \int \frac{4}{x ^{3} + 4 x} d x

\int \frac{6 x ^{2}}{x ^{3} + 4 x} d x = 3 ln x^{2} + 4

\int \frac{5 x}{x ^{3} + 4 x} d x = \frac{5}{2} arctan (\frac{x}{2})

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

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

Simplify

3 ln x^{2} + 4 - \frac{5}{2} arctan (\frac{x}{2}) + ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 4 : - \frac{5}{2} arctan (\frac{x}{2}) + \frac{5}{2} ln x^{2} + 4 + ln ∣ x ∣

= - \frac{5}{2} arctan (\frac{x}{2}) + \frac{5}{2} ln x^{2} + 4 + ln ∣ x ∣

Add a constant to the solution

= - \frac{5}{2} arctan (\frac{x}{2}) + \frac{5}{2} ln x^{2} + 4 + ln ∣ x ∣ + C

x \to 0 + lim (cot (x) - \frac{1}{x})

\int 1 + 4 x d x

\frac{1}{s} \cdot \frac{1}{s + 1}

\int 8 sin^{2} (x + \frac{π}{12}) d x

n = 1 \sum \infty \infty

What is the integral of (6x^2-5x+4)/(x^3+4x) ?
The integral of (6x^2-5x+4)/(x^3+4x) is -5/2 arctan(x/2)+5/2 ln|x^2+4|+ln|x|+C