What is the integral of 5/(x(x^4+9)) ?

The integral of 5/(x(x^4+9)) is 5/9 ln(x)-5/36 ln|x^4+9|+C

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integral of 5/(x(x^4+9))

\int \frac{5}{x ( x ^{4} + 9 )} d x

\frac{5}{9} ln (x) - \frac{5}{36} ln x^{4} + 9 + C

Solution steps

\int \frac{5}{x ( x ^{4} + 9 )} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{x ( x ^{4} + 9 )} d x

Apply u-substitution

= 5 \cdot \int \frac{1}{4 u ( u + 9 )} d u

Take the constant out:

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

= 5 \cdot \frac{1}{4} \cdot \int \frac{1}{u ( u + 9 )} d u

Take the partial fraction of

\frac{1}{u ( u + 9 )} : \frac{1}{9 u} - \frac{1}{9 ( u + 9 )}

= 5 \cdot \frac{1}{4} \cdot \int \frac{1}{9 u} - \frac{1}{9 ( u + 9 )} 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

= 5 \cdot \frac{1}{4} (\int \frac{1}{9 u} d u - \int \frac{1}{9 ( u + 9 )} d u)

\int \frac{1}{9 u} d u = \frac{1}{9} ln ∣ u ∣

\int \frac{1}{9 ( u + 9 )} d u = \frac{1}{9} ln ∣ u + 9 ∣

= 5 \cdot \frac{1}{4} (\frac{1}{9} ln ∣ u ∣ - \frac{1}{9} ln ∣ u + 9 ∣)

Substitute back

u = x^{4}

= 5 \cdot \frac{1}{4} (\frac{1}{9} ln x^{4} - \frac{1}{9} ln x^{4} + 9)

Simplify

5 \cdot \frac{1}{4} (\frac{1}{9} ln x^{4} - \frac{1}{9} ln x^{4} + 9) : \frac{5}{9} ln (x) - \frac{5}{36} ln x^{4} + 9

= \frac{5}{9} ln (x) - \frac{5}{36} ln x^{4} + 9

Add a constant to the solution

= \frac{5}{9} ln (x) - \frac{5}{36} ln x^{4} + 9 + C

What is the integral of 5/(x(x^4+9)) ?
The integral of 5/(x(x^4+9)) is 5/9 ln(x)-5/36 ln|x^4+9|+C