What is the integral of (x^2-4x+1)/(x^2+9) ?

The integral of (x^2-4x+1)/(x^2+9) is x-8/3 arctan(x/3)-2ln|x^2+9|+C

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integral of (x^2-4x+1)/(x^2+9)

\int \frac{x ^{2} - 4 x + 1}{x ^{2} + 9} d x

x - \frac{8}{3} arctan (\frac{x}{3}) - 2 ln x^{2} + 9 + C

Solution steps

\int \frac{x ^{2} - 4 x + 1}{x ^{2} + 9} d x

Expand

\frac{x ^{2} - 4 x + 1}{x ^{2} + 9} : \frac{x ^{2}}{x ^{2} + 9} - \frac{4 x}{x ^{2} + 9} + \frac{1}{x ^{2} + 9}

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{x ^{2}}{x ^{2} + 9} d x - \int \frac{4 x}{x ^{2} + 9} d x + \int \frac{1}{x ^{2} + 9} d x

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

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

\int \frac{1}{x ^{2} + 9} d x = \frac{1}{3} arctan (\frac{x}{3})

= - 3 arctan (\frac{x}{3}) + x - 2 ln x^{2} + 9 + \frac{1}{3} arctan (\frac{x}{3})

Simplify

- 3 arctan (\frac{x}{3}) + x - 2 ln x^{2} + 9 + \frac{1}{3} arctan (\frac{x}{3}) : x - \frac{8}{3} arctan (\frac{x}{3}) - 2 ln x^{2} + 9

= x - \frac{8}{3} arctan (\frac{x}{3}) - 2 ln x^{2} + 9

Add a constant to the solution

= x - \frac{8}{3} arctan (\frac{x}{3}) - 2 ln x^{2} + 9 + C

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What is the integral of (x^2-4x+1)/(x^2+9) ?
The integral of (x^2-4x+1)/(x^2+9) is x-8/3 arctan(x/3)-2ln|x^2+9|+C