What is the integral of (-8x+5)/(x^3+4x) ?

The integral of (-8x+5)/(x^3+4x) is 5/4 ln|x|-1/4 (5/2 ln|(x^2)/4+1|+16arctan(x/2))+C

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integral of (-8x+5)/(x^3+4x)

\int \frac{- 8 x + 5}{x ^{3} + 4 x} d x

\frac{5}{4} ln ∣ x ∣ - \frac{1}{4} (\frac{5}{2} ln \frac{x ^{2}}{4} + 1 + 16 arctan (\frac{x}{2})) + C

Solution steps

\int \frac{- 8 x + 5}{x ^{3} + 4 x} d x

Take the partial fraction of

\frac{- 8 x + 5}{x ^{3} + 4 x} : \frac{5}{4 x} + \frac{- 5 x - 32}{4 ( x ^{2} + 4 )}

= \int \frac{5}{4 x} + \frac{- 5 x - 32}{4 ( x ^{2} + 4 )} 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 \frac{5}{4 x} d x + \int \frac{- 5 x - 32}{4 ( x ^{2} + 4 )} d x

\int \frac{5}{4 x} d x = \frac{5}{4} ln ∣ x ∣

\int \frac{- 5 x - 32}{4 ( x ^{2} + 4 )} d x = - \frac{1}{4} (\frac{5}{2} ln \frac{x ^{2}}{4} + 1 + 16 arctan (\frac{x}{2}))

= \frac{5}{4} ln ∣ x ∣ - \frac{1}{4} (\frac{5}{2} ln \frac{x ^{2}}{4} + 1 + 16 arctan (\frac{x}{2}))

Add a constant to the solution

= \frac{5}{4} ln ∣ x ∣ - \frac{1}{4} (\frac{5}{2} ln \frac{x ^{2}}{4} + 1 + 16 arctan (\frac{x}{2})) + C

x \to 36 lim (x^{\frac{3}{2}})

y^{^{''}} + x = 7 y

\frac{\partial}{\partial y} (2 x e^{- 2 x y})

\frac{\partial}{\partial x} (z^{2} x)

x + x + x

What is the integral of (-8x+5)/(x^3+4x) ?
The integral of (-8x+5)/(x^3+4x) is 5/4 ln|x|-1/4 (5/2 ln|(x^2)/4+1|+16arctan(x/2))+C