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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{4 u ( u + 4 )} d u

= \int \frac{1}{4 u ( u + 4 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int \frac{1}{u ( u + 4 )} d u

Take the partial fraction of

\frac{1}{u ( u + 4 )} : \frac{1}{4 u} - \frac{1}{4 ( u + 4 )}

= \frac{1}{4} \cdot \int \frac{1}{4 u} - \frac{1}{4 ( u + 4 )} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{4} (\int \frac{1}{4 u} d u - \int \frac{1}{4 ( u + 4 )} d u)

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

\int \frac{1}{4 ( u + 4 )} d u = \frac{1}{4} ln ∣ u + 4 ∣

= \frac{1}{4} (\frac{1}{4} ln ∣ u ∣ - \frac{1}{4} ln ∣ u + 4 ∣)

Substitute back

u = x^{4}

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

\frac{1}{4} ln x^{4} = ln (x)

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

Add a constant to the solution

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

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