What is the integral of (81)/(x^4+27x) ?

The integral of (81)/(x^4+27x) is 3ln|x|-ln|x+3|-ln|4x^2-12x+36|+C

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integral of (81)/(x^4+27x)

\int \frac{81}{x ^{4} + 27 x} d x

3 ln ∣ x ∣ - ln ∣ x + 3 ∣ - ln 4 x^{2} - 12 x + 36 + C

Solution steps

\int \frac{81}{x ^{4} + 27 x} d x

Take the constant out:

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

= 81 \cdot \int \frac{1}{x ^{4} + 27 x} d x

Take the partial fraction of

\frac{1}{x ^{4} + 27 x} : \frac{1}{27 x} - \frac{1}{81 ( x + 3 )} + \frac{- 2 x + 3}{81 ( x ^{2} - 3 x + 9 )}

= 81 \cdot \int \frac{1}{27 x} - \frac{1}{81 ( x + 3 )} + \frac{- 2 x + 3}{81 ( x ^{2} - 3 x + 9 )} d x

Apply the Sum Rule:

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

= 81 (\int \frac{1}{27 x} d x - \int \frac{1}{81 ( x + 3 )} d x + \int \frac{- 2 x + 3}{81 ( x ^{2} - 3 x + 9 )} d x)

\int \frac{1}{27 x} d x = \frac{1}{27} ln ∣ x ∣

\int \frac{1}{81 ( x + 3 )} d x = \frac{1}{81} ln ∣ x + 3 ∣

\int \frac{- 2 x + 3}{81 ( x ^{2} - 3 x + 9 )} d x = - \frac{1}{81} ln 4 x^{2} - 12 x + 36

= 81 (\frac{1}{27} ln ∣ x ∣ - \frac{1}{81} ln ∣ x + 3 ∣ - \frac{1}{81} ln 4 x^{2} - 12 x + 36)

Simplify

81 (\frac{1}{27} ln ∣ x ∣ - \frac{1}{81} ln ∣ x + 3 ∣ - \frac{1}{81} ln 4 x^{2} - 12 x + 36) : 3 ln ∣ x ∣ - ln ∣ x + 3 ∣ - ln 4 x^{2} - 12 x + 36

= 3 ln ∣ x ∣ - ln ∣ x + 3 ∣ - ln 4 x^{2} - 12 x + 36

Add a constant to the solution

= 3 ln ∣ x ∣ - ln ∣ x + 3 ∣ - ln 4 x^{2} - 12 x + 36 + C

\frac{d}{d x} (- 7 x^{2} + 400 x)

y^{^{''}} - 4 y = 8 e^{2 x}

x \to 3 lim (x + 12)

\int_{0}^{2 π} sin (θ) d θ

\frac{d y}{d x} = e^{x + y + 3}

What is the integral of (81)/(x^4+27x) ?
The integral of (81)/(x^4+27x) is 3ln|x|-ln|x+3|-ln|4x^2-12x+36|+C