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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

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

Add a constant to the solution

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

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

\frac{d}{d x} (- x^{2} + 25)

y d x + x^{\frac{α}{6}} d y = 8 y x d x + 2 d y

t a y l or ln (x^{2} + 1)

d er i v a t i v e - 6 t (5 t^{3} - 1)^{7}

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