What is the integral of (10x^2-9x+8)/(x^3+4x) ?

The integral of (10x^2-9x+8)/(x^3+4x) is -9/2 arctan(x/2)+4ln|x^2+4|+2ln|x|+C

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integral of (10x^2-9x+8)/(x^3+4x)

\int \frac{10 x ^{2} - 9 x + 8}{x ^{3} + 4 x} d x

- \frac{9}{2} arctan (\frac{x}{2}) + 4 ln x^{2} + 4 + 2 ln ∣ x ∣ + C

Solution steps

\int \frac{10 x ^{2} - 9 x + 8}{x ^{3} + 4 x} d x

Expand

\frac{10 x ^{2} - 9 x + 8}{x ^{3} + 4 x} : \frac{10 x ^{2}}{x ^{3} + 4 x} - \frac{9 x}{x ^{3} + 4 x} + \frac{8}{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{10 x ^{2}}{x ^{3} + 4 x} d x - \int \frac{9 x}{x ^{3} + 4 x} d x + \int \frac{8}{x ^{3} + 4 x} d x

\int \frac{10 x ^{2}}{x ^{3} + 4 x} d x = 5 ln x^{2} + 4

\int \frac{9 x}{x ^{3} + 4 x} d x = \frac{9}{2} arctan (\frac{x}{2})

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

= 5 ln x^{2} + 4 - \frac{9}{2} arctan (\frac{x}{2}) + 2 ln ∣ x ∣ - ln x^{2} + 4

Simplify

= - \frac{9}{2} arctan (\frac{x}{2}) + 4 ln x^{2} + 4 + 2 ln ∣ x ∣

Add a constant to the solution

= - \frac{9}{2} arctan (\frac{x}{2}) + 4 ln x^{2} + 4 + 2 ln ∣ x ∣ + C

f (x) = 19 x - 1.86 x^{2}, at x = 1

\frac{\partial}{\partial x} (\frac{10}{x})

f (x) = \frac{2 x}{x + 1}, at x = 1

\frac{d y}{d x} + 4 x y^{2} = 0

\int \frac{4 x ^{3}}{x ^{4} + 1} d x

What is the integral of (10x^2-9x+8)/(x^3+4x) ?
The integral of (10x^2-9x+8)/(x^3+4x) is -9/2 arctan(x/2)+4ln|x^2+4|+2ln|x|+C