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

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

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

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

\frac{1}{2} arctan (\frac{x}{2}) - \frac{1}{4} ln ∣ x ∣ + \frac{1}{8} ln x^{2} + 4 + C

Solution steps

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

Expand

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

\int \frac{x}{x ^{3} + 4 x} d x = \frac{1}{2} arctan (\frac{x}{2})

\int \frac{1}{x ^{3} + 4 x} d x = \frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4

= \frac{1}{2} arctan (\frac{x}{2}) - (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

Simplify

= \frac{1}{2} arctan (\frac{x}{2}) - \frac{1}{4} ln ∣ x ∣ + \frac{1}{8} ln x^{2} + 4

Add a constant to the solution

= \frac{1}{2} arctan (\frac{x}{2}) - \frac{1}{4} ln ∣ x ∣ + \frac{1}{8} ln x^{2} + 4 + C

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