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

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

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

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

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

Solution steps

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

Long division

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Add a constant to the solution

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

\frac{d y}{d x} + \frac{y}{x} = \frac{1}{x}

sin^{3} (4 x)

y^{^{'}} - 5 y = 0

\frac{d}{d x} (\frac{1}{( 1 - 2500 x ) ^{4}})

\int \frac{1}{t ^{4}} d t

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