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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

\int \frac{x}{4 ( x ^{2} + 4 )} d x = \frac{1}{8} ln x^{2} + 4

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

Add a constant to the solution

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

f (x) = 4 x^{\frac{5}{4}} + 8 x^{\frac{3}{2}} + 2 x

f (x) = \frac{2}{x}, at \frac{1}{16}

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

\int (x^{5} - 10 x^{4} + x^{3} - 15 x^{2}) d x

n = 1 \sum \infty \frac{z ^{n}}{n ^{2}}

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