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

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

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

\int \frac{( 2 x ^{2} - x + 4 )}{x ^{3} + 4 x} d x

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

Solution steps

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

Take the partial fraction of

\frac{2 x ^{2} - x + 4}{x ^{3} + 4 x} : \frac{1}{x} + \frac{x - 1}{x ^{2} + 4}

= \int \frac{1}{x} + \frac{x - 1}{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{1}{x} d x + \int \frac{x - 1}{x ^{2} + 4} d x

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

Add a constant to the solution

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

x \to 2 lim (\frac{x ^{3} + x ^{2} - 8 x - 12}{x ^{4} + 4 x ^{3} + 5 x ^{2} + 4 x + 4})

\frac{\partial}{\partial x} (z^{3} - x y + yz + y^{3} - 2)

\frac{4}{[ s ( s ^{2} + 4 ) ]}

y^{^{'}} = \frac{( t ^{4} + y ^{4} )}{t y ^{3}}

\int \frac{3 - x}{x ^{2} - 1} d x

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