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

The integral of (x^4+2)/(x^3+9x) is (x^2)/2-83/18 ln|x^2+9|+2/9 ln|x|+C

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

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

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

Solution steps

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

Long division

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

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

\int x d x = \frac{x ^{2}}{2}

\int \frac{- 9 x ^{2} + 2}{x ^{3} + 9 x} d x = - \frac{83}{18} ln x^{2} + 9 + \frac{2}{9} ln ∣ x ∣

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

Add a constant to the solution

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

\int \frac{x + 1}{x ^{5}} d x

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

\frac{d y}{d x}, y = x^{2} + 5

- 1.8 t^{2} + 14.4 t

y = t sin (t) + cos (t)

What is the integral of (x^4+2)/(x^3+9x) ?
The integral of (x^4+2)/(x^3+9x) is (x^2)/2-83/18 ln|x^2+9|+2/9 ln|x|+C