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

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

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

Expand

\frac{10 x ^{2} - 9 x + 4}{x ( x ^{2} + 4 )} : \frac{10 x ^{2}}{x ( x ^{2} + 4 )} - \frac{9 x}{x ( x ^{2} + 4 )} + \frac{4}{x ( x ^{2} + 4 )}

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 ( x ^{2} + 4 )} d x - \int \frac{9 x}{x ( x ^{2} + 4 )} d x + \int \frac{4}{x ( x ^{2} + 4 )} d x

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

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

\int \frac{4}{x ( x ^{2} + 4 )} 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

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