What is the integral of (-9x^2+2)/(x^3+9x) ?

The integral of (-9x^2+2)/(x^3+9x) is -83/18 ln|x^2+9|+2/9 ln|x|+C

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integral of (-9x^2+2)/(x^3+9x)

\int \frac{- 9 x ^{2} + 2}{x ^{3} + 9 x} d x

- \frac{83}{18} ln x^{2} + 9 + \frac{2}{9} ln ∣ x ∣ + C

Solution steps

\int \frac{- 9 x ^{2} + 2}{x ^{3} + 9 x} d x

Expand

\frac{- 9 x ^{2} + 2}{x ^{3} + 9 x} : - \frac{9 x ^{2}}{x ^{3} + 9 x} + \frac{2}{x ^{3} + 9 x}

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

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

\int \frac{2}{x ^{3} + 9 x} d x = 2 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln x^{2} + 9)

= - \frac{9}{2} ln x^{2} + 9 + 2 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln x^{2} + 9)

Simplify

- \frac{9}{2} ln x^{2} + 9 + 2 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln x^{2} + 9) : - \frac{83}{18} ln x^{2} + 9 + \frac{2}{9} ln ∣ x ∣

= - \frac{83}{18} ln x^{2} + 9 + \frac{2}{9} ln ∣ x ∣

Add a constant to the solution

= - \frac{83}{18} ln x^{2} + 9 + \frac{2}{9} ln ∣ x ∣ + C

\int \frac{1}{x ^{2} x ^{2} + 49} d x

\frac{d}{d x} (\frac{2}{( 1 - \frac{1}{3} x ^{3} + 1 ) ^{2}})

5 x^{2} - 2 x - 1

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

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{- 2 x} ln (x)

What is the integral of (-9x^2+2)/(x^3+9x) ?
The integral of (-9x^2+2)/(x^3+9x) is -83/18 ln|x^2+9|+2/9 ln|x|+C