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

The integral of (9x^2-8x+8)/(x(x^2+4)) is 7/2 ln|x^2+4|-4arctan(x/2)+2ln|x|+C

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

\int \frac{9 x ^{2} - 8 x + 8}{x ( x ^{2} + 4 )} d x

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

Solution steps

\int \frac{9 x ^{2} - 8 x + 8}{x ( x ^{2} + 4 )} d x

Expand

\frac{9 x ^{2} - 8 x + 8}{x ( x ^{2} + 4 )} : \frac{9 x ^{2}}{x ( x ^{2} + 4 )} - \frac{8 x}{x ( x ^{2} + 4 )} + \frac{8}{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{9 x ^{2}}{x ( x ^{2} + 4 )} d x - \int \frac{8 x}{x ( x ^{2} + 4 )} d x + \int \frac{8}{x ( x ^{2} + 4 )} d x

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

\int \frac{8 x}{x ( x ^{2} + 4 )} d x = 4 arctan (\frac{x}{2})

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

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

Add similar elements:

\frac{9}{2} ln x^{2} + 4 - ln x^{2} + 4 = \frac{7}{2} ln x^{2} + 4

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

Add a constant to the solution

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

n = 0 \sum \infty \frac{n}{n + 2}

\frac{d}{d x} (\frac{x}{1 - 2 x})

\int x^{3} x^{2} + 15 d x

n = 1 \sum \infty \frac{1}{2 ^{n} ( n + 1 )}

x \to 9 lim (\frac{4 x}{x - 9})

What is the integral of (9x^2-8x+8)/(x(x^2+4)) ?
The integral of (9x^2-8x+8)/(x(x^2+4)) is 7/2 ln|x^2+4|-4arctan(x/2)+2ln|x|+C