What is the integral of (5x^3+7x+2)/(x^4+9x^2) ?

The integral of (5x^3+7x+2)/(x^4+9x^2) is 19/9 ln|x^2+9|+7/9 ln(x)-2/27 arctan(x/3)-2/(9x)+C

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integral of (5x^3+7x+2)/(x^4+9x^2)

\int \frac{5 x ^{3} + 7 x + 2}{x ^{4} + 9 x ^{2}} d x

\frac{19}{9} ln x^{2} + 9 + \frac{7}{9} ln (x) - \frac{2}{27} arctan (\frac{x}{3}) - \frac{2}{9 x} + C

Solution steps

\int \frac{5 x ^{3} + 7 x + 2}{x ^{4} + 9 x ^{2}} d x

Expand

\frac{5 x ^{3} + 7 x + 2}{x ^{4} + 9 x ^{2}} : \frac{5 x ^{3}}{x ^{4} + 9 x ^{2}} + \frac{7 x}{x ^{4} + 9 x ^{2}} + \frac{2}{x ^{4} + 9 x ^{2}}

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{5 x ^{3}}{x ^{4} + 9 x ^{2}} d x + \int \frac{7 x}{x ^{4} + 9 x ^{2}} d x + \int \frac{2}{x ^{4} + 9 x ^{2}} d x

\int \frac{5 x ^{3}}{x ^{4} + 9 x ^{2}} d x = \frac{5}{2} ln x^{2} + 9

\int \frac{7 x}{x ^{4} + 9 x ^{2}} d x = 7 (\frac{1}{9} ln (x) - \frac{1}{18} ln x^{2} + 9)

\int \frac{2}{x ^{4} + 9 x ^{2}} d x = 2 (- \frac{1}{9 x} - \frac{1}{27} arctan (\frac{x}{3}))

= \frac{5}{2} ln x^{2} + 9 + 7 (\frac{1}{9} ln (x) - \frac{1}{18} ln x^{2} + 9) + 2 (- \frac{1}{9 x} - \frac{1}{27} arctan (\frac{x}{3}))

Simplify

\frac{5}{2} ln x^{2} + 9 + 7 (\frac{1}{9} ln (x) - \frac{1}{18} ln x^{2} + 9) + 2 (- \frac{1}{9 x} - \frac{1}{27} arctan (\frac{x}{3})) : \frac{19}{9} ln x^{2} + 9 + \frac{7}{9} ln (x) - \frac{2}{27} arctan (\frac{x}{3}) - \frac{2}{9 x}

= \frac{19}{9} ln x^{2} + 9 + \frac{7}{9} ln (x) - \frac{2}{27} arctan (\frac{x}{3}) - \frac{2}{9 x}

Add a constant to the solution

= \frac{19}{9} ln x^{2} + 9 + \frac{7}{9} ln (x) - \frac{2}{27} arctan (\frac{x}{3}) - \frac{2}{9 x} + C

What is the integral of (5x^3+7x+2)/(x^4+9x^2) ?
The integral of (5x^3+7x+2)/(x^4+9x^2) is 19/9 ln|x^2+9|+7/9 ln(x)-2/27 arctan(x/3)-2/(9x)+C