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

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

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

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

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

Solution steps

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

Expand

\frac{10 x ^{2} - 9 x + 4}{x ^{3} + 4 x} : \frac{10 x ^{2}}{x ^{3} + 4 x} - \frac{9 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{10 x ^{2}}{x ^{3} + 4 x} d x - \int \frac{9 x}{x ^{3} + 4 x} d x + \int \frac{4}{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{4}{x ^{3} + 4 x} d x = ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 4

= 5 ln x^{2} + 4 - \frac{9}{2} arctan (\frac{x}{2}) + ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 4

Simplify

5 ln x^{2} + 4 - \frac{9}{2} arctan (\frac{x}{2}) + ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 4 : - \frac{9}{2} arctan (\frac{x}{2}) + \frac{9}{2} ln x^{2} + 4 + ln ∣ x ∣

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{7 x ^{2}}{( x ^{2} + x ) ^{3}})

n = 0 \sum \infty \frac{x ^{n}}{3 + 2 x}

\frac{\partial}{\partial x} (ln (x^{2} + y^{2} + 3))

4 sin (x)

\frac{5}{s + 5}

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