What is the integral of 9/(x(x^4+8)) ?

The integral of 9/(x(x^4+8)) is 9/4 (1/2 ln(x)-1/8 ln|x^4+8|)+C

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integral of 9/(x(x^4+8))

\int \frac{9}{x ( x ^{4} + 8 )} d x

\frac{9}{4} (\frac{1}{2} ln (x) - \frac{1}{8} ln x^{4} + 8) + C

Solution steps

\int \frac{9}{x ( x ^{4} + 8 )} 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} + 8 )} d x

Apply u-substitution

: \int \frac{1}{4 u ( u + 8 )} d u

= 9 \cdot \int \frac{1}{4 u ( u + 8 )} 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 + 8 )} d u

Take the partial fraction of

\frac{1}{u ( u + 8 )} : \frac{1}{8 u} - \frac{1}{8 ( u + 8 )}

= 9 \cdot \frac{1}{4} \cdot \int \frac{1}{8 u} - \frac{1}{8 ( u + 8 )} 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}{8 u} d u - \int \frac{1}{8 ( u + 8 )} d u)

\int \frac{1}{8 u} d u = \frac{1}{8} ln ∣ u ∣

\int \frac{1}{8 ( u + 8 )} d u = \frac{1}{8} ln ∣ u + 8 ∣

= 9 \cdot \frac{1}{4} (\frac{1}{8} ln ∣ u ∣ - \frac{1}{8} ln ∣ u + 8 ∣)

Substitute back

u = x^{4}

= 9 \cdot \frac{1}{4} (\frac{1}{8} ln x^{4} - \frac{1}{8} ln x^{4} + 8)

Simplify

9 \cdot \frac{1}{4} (\frac{1}{8} ln x^{4} - \frac{1}{8} ln x^{4} + 8) : \frac{9}{4} (\frac{1}{2} ln (x) - \frac{1}{8} ln x^{4} + 8)

= \frac{9}{4} (\frac{1}{2} ln (x) - \frac{1}{8} ln x^{4} + 8)

Add a constant to the solution

= \frac{9}{4} (\frac{1}{2} ln (x) - \frac{1}{8} ln x^{4} + 8) + C

y = \frac{5 x}{x + 4}, (1, 1)

\int \frac{cos ( x )}{sin ( 2 x )} d x

\frac{1}{x ^{2}}, 2

(2 x^{2} y + 2 1 + x^{4} y^{2}) d x + x^{3} d y = 0

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

What is the integral of 9/(x(x^4+8)) ?
The integral of 9/(x(x^4+8)) is 9/4 (1/2 ln(x)-1/8 ln|x^4+8|)+C