What is the integral of (x+1)/(x^3+4x) ?

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

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integral of (x+1)/(x^3+4x)

\int \frac{x + 1}{x ^{3} + 4 x} d x

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

Solution steps

\int \frac{x + 1}{x ^{3} + 4 x} d x

Take the partial fraction of

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

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

\int \frac{1}{4 x} d x = \frac{1}{4} ln ∣ x ∣

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

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

Add a constant to the solution

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

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