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

The integral of 9/(x(x^4+5)) is 9/5 ln(x)-9/20 ln|x^4+5|+C

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

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

\frac{9}{5} ln (x) - \frac{9}{20} ln x^{4} + 5 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{4 u ( u + 5 )} d u

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

Take the partial fraction of

\frac{1}{u ( u + 5 )} : \frac{1}{5 u} - \frac{1}{5 ( u + 5 )}

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

\int \frac{1}{5 u} d u = \frac{1}{5} ln ∣ u ∣

\int \frac{1}{5 ( u + 5 )} d u = \frac{1}{5} ln ∣ u + 5 ∣

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

Substitute back

u = x^{4}

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

Simplify

9 \cdot \frac{1}{4} (\frac{1}{5} ln x^{4} - \frac{1}{5} ln x^{4} + 5) : \frac{9}{5} ln (x) - \frac{9}{20} ln x^{4} + 5

= \frac{9}{5} ln (x) - \frac{9}{20} ln x^{4} + 5

Add a constant to the solution

= \frac{9}{5} ln (x) - \frac{9}{20} ln x^{4} + 5 + C

f (t) = 2 cos (t) + sin (2 t)

f (x) = x^{2} - 1

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

\frac{d}{d x} (ln (x - 7))

\frac{10 x + x ^{2}}{( 5 + x ) ^{2}}

What is the integral of 9/(x(x^4+5)) ?
The integral of 9/(x(x^4+5)) is 9/5 ln(x)-9/20 ln|x^4+5|+C