What is the integral of 9/(x(x^4+2)) ?

The integral of 9/(x(x^4+2)) is 9/4 (2ln(x)-1/2 ln|x^4+2|)+C

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integral of 9/(x(x^4+2))

\int \frac{9}{x ( x ^{4} + 2 )} d x

\frac{9}{4} (2 ln (x) - \frac{1}{2} ln x^{4} + 2) + C

Solution steps

\int \frac{9}{x ( x ^{4} + 2 )} d x

Take the constant out:

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

= 9 \cdot \int \frac{1}{x ( x ^{4} + 2 )} d x

Apply u-substitution

: \int \frac{1}{4 u ( u + 2 )} d u

= 9 \cdot \int \frac{1}{4 u ( u + 2 )} d u

Take the constant out:

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

= 9 \cdot \frac{1}{4} \cdot \int \frac{1}{u ( u + 2 )} d u

Take the partial fraction of

\frac{1}{u ( u + 2 )} : \frac{1}{2 u} - \frac{1}{2 ( u + 2 )}

= 9 \cdot \frac{1}{4} \cdot \int \frac{1}{2 u} - \frac{1}{2 ( u + 2 )} 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

= 9 \cdot \frac{1}{4} (\int \frac{1}{2 u} d u - \int \frac{1}{2 ( u + 2 )} d u)

\int \frac{1}{2 u} d u = \frac{1}{2} ln ∣ u ∣

\int \frac{1}{2 ( u + 2 )} d u = \frac{1}{2} ln ∣ u + 2 ∣

= 9 \cdot \frac{1}{4} (\frac{1}{2} ln ∣ u ∣ - \frac{1}{2} ln ∣ u + 2 ∣)

Substitute back

u = x^{4}

= 9 \cdot \frac{1}{4} (\frac{1}{2} ln x^{4} - \frac{1}{2} ln x^{4} + 2)

Simplify

9 \cdot \frac{1}{4} (\frac{1}{2} ln x^{4} - \frac{1}{2} ln x^{4} + 2) : \frac{9}{4} (2 ln (x) - \frac{1}{2} ln x^{4} + 2)

= \frac{9}{4} (2 ln (x) - \frac{1}{2} ln x^{4} + 2)

Add a constant to the solution

= \frac{9}{4} (2 ln (x) - \frac{1}{2} ln x^{4} + 2) + C

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What is the integral of 9/(x(x^4+2)) ?
The integral of 9/(x(x^4+2)) is 9/4 (2ln(x)-1/2 ln|x^4+2|)+C