What is the integral of (8x^2-7x+12)/(x^3+4x) ?

The integral of (8x^2-7x+12)/(x^3+4x) is -7/2 arctan(x/2)+5/2 ln|x^2+4|+3ln|x|+C

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integral of (8x^2-7x+12)/(x^3+4x)

\int \frac{8 x ^{2} - 7 x + 12}{x ^{3} + 4 x} d x

- \frac{7}{2} arctan (\frac{x}{2}) + \frac{5}{2} ln x^{2} + 4 + 3 ln ∣ x ∣ + C

Solution steps

\int \frac{8 x ^{2} - 7 x + 12}{x ^{3} + 4 x} d x

Expand

\frac{8 x ^{2} - 7 x + 12}{x ^{3} + 4 x} : \frac{8 x ^{2}}{x ^{3} + 4 x} - \frac{7 x}{x ^{3} + 4 x} + \frac{12}{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{8 x ^{2}}{x ^{3} + 4 x} d x - \int \frac{7 x}{x ^{3} + 4 x} d x + \int \frac{12}{x ^{3} + 4 x} d x

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

\int \frac{7 x}{x ^{3} + 4 x} d x = \frac{7}{2} arctan (\frac{x}{2})

\int \frac{12}{x ^{3} + 4 x} d x = 12 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

= 4 ln x^{2} + 4 - \frac{7}{2} arctan (\frac{x}{2}) + 12 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

Simplify

4 ln x^{2} + 4 - \frac{7}{2} arctan (\frac{x}{2}) + 12 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4) : - \frac{7}{2} arctan (\frac{x}{2}) + \frac{5}{2} ln x^{2} + 4 + 3 ln ∣ x ∣

= - \frac{7}{2} arctan (\frac{x}{2}) + \frac{5}{2} ln x^{2} + 4 + 3 ln ∣ x ∣

Add a constant to the solution

= - \frac{7}{2} arctan (\frac{x}{2}) + \frac{5}{2} ln x^{2} + 4 + 3 ln ∣ x ∣ + C

\frac{d}{d x} ((- 2 x + 3 x^{- 2}) ln (x))

\frac{d}{d x} (\frac{x ^{7}}{x ^{4}})

x \to 0 lim (1 + \frac{1}{x ^{2}})

x^{3} - 2 x^{2} y - 2 y^{2} - 1

arcsin (\frac{x}{7})

What is the integral of (8x^2-7x+12)/(x^3+4x) ?
The integral of (8x^2-7x+12)/(x^3+4x) is -7/2 arctan(x/2)+5/2 ln|x^2+4|+3ln|x|+C