What is the integral of (x^2-16x+9)/(x^4+10x^2+9) ?

The integral of (x^2-16x+9)/(x^4+10x^2+9) is arctan(x)-ln|x^2+1|+ln|x^2+9|+C

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integral of (x^2-16x+9)/(x^4+10x^2+9)

\int \frac{x ^{2} - 16 x + 9}{x ^{4} + 10 x ^{2} + 9} d x

arctan (x) - ln x^{2} + 1 + ln x^{2} + 9 + C

Solution steps

\int \frac{x ^{2} - 16 x + 9}{x ^{4} + 10 x ^{2} + 9} d x

Expand

\frac{x ^{2} - 16 x + 9}{x ^{4} + 10 x ^{2} + 9} : \frac{x ^{2}}{x ^{4} + 10 x ^{2} + 9} - \frac{16 x}{x ^{4} + 10 x ^{2} + 9} + \frac{9}{x ^{4} + 10 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 ^{4} + 10 x ^{2} + 9} d x - \int \frac{16 x}{x ^{4} + 10 x ^{2} + 9} d x + \int \frac{9}{x ^{4} + 10 x ^{2} + 9} d x

\int \frac{x ^{2}}{x ^{4} + 10 x ^{2} + 9} d x = - \frac{1}{8} arctan (x) + \frac{3}{8} arctan (\frac{x}{3})

\int \frac{16 x}{x ^{4} + 10 x ^{2} + 9} d x = ln x^{2} + 1 - ln x^{2} + 9

\int \frac{9}{x ^{4} + 10 x ^{2} + 9} d x = 9 (\frac{1}{8} arctan (x) - \frac{1}{24} arctan (\frac{x}{3}))

= - \frac{1}{8} arctan (x) + \frac{3}{8} arctan (\frac{x}{3}) - (ln x^{2} + 1 - ln x^{2} + 9) + 9 (\frac{1}{8} arctan (x) - \frac{1}{24} arctan (\frac{x}{3}))

Simplify

- \frac{1}{8} arctan (x) + \frac{3}{8} arctan (\frac{x}{3}) - (ln x^{2} + 1 - ln x^{2} + 9) + 9 (\frac{1}{8} arctan (x) - \frac{1}{24} arctan (\frac{x}{3})) : arctan (x) - ln x^{2} + 1 + ln x^{2} + 9

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

Add a constant to the solution

= arctan (x) - ln x^{2} + 1 + ln x^{2} + 9 + C

y = \frac{3}{2} x

x \to 0 lim (\frac{2 ^{x}}{x})

x \to 0 + lim (9 sin (x) ln (x))

x \to 0 - lim (- \frac{1}{3 x ^{3}})

\int_{0}^{\frac{π}{9}} 7 - 7 cos (3 x) d x

What is the integral of (x^2-16x+9)/(x^4+10x^2+9) ?
The integral of (x^2-16x+9)/(x^4+10x^2+9) is arctan(x)-ln|x^2+1|+ln|x^2+9|+C