What is the integral of (3x^39x^2-15)/(x^24x^4) ?

The integral of (3x^39x^2-15)/(x^24x^4) is 1/4 (27ln|x|+3/(x^5))+C

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integral of (3x^39x^2-15)/(x^24x^4)

\int \frac{3 x ^{3} 9 x ^{2} - 15}{x ^{2} 4 x ^{4}} d x

\frac{1}{4} (27 ln ∣ x ∣ + \frac{3}{x ^{5}}) + C

Solution steps

\int \frac{3 x ^{3} \cdot 9 x ^{2} - 15}{x ^{2} \cdot 4 x ^{4}} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{3 x ^{3} \cdot 9 x ^{2} - 15}{x ^{2} x ^{4}} d x

Simplify

\frac{3 x ^{3} \cdot 9 x ^{2} - 15}{x ^{2} x ^{4}} : \frac{27 x ^{5} - 15}{x ^{6}}

= \frac{1}{4} \cdot \int \frac{27 x ^{5} - 15}{x ^{6}} d x

Expand

\frac{27 x ^{5} - 15}{x ^{6}} : \frac{27}{x} - \frac{15}{x ^{6}}

= \frac{1}{4} \cdot \int \frac{27}{x} - \frac{15}{x ^{6}} 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

= \frac{1}{4} (\int \frac{27}{x} d x - \int \frac{15}{x ^{6}} d x)

\int \frac{27}{x} d x = 27 ln ∣ x ∣

\int \frac{15}{x ^{6}} d x = - \frac{3}{x ^{5}}

= \frac{1}{4} (27 ln ∣ x ∣ - (- \frac{3}{x ^{5}}))

Simplify

= \frac{1}{4} (27 ln ∣ x ∣ + \frac{3}{x ^{5}})

Add a constant to the solution

= \frac{1}{4} (27 ln ∣ x ∣ + \frac{3}{x ^{5}}) + C

\frac{d}{d x} (arcsin (3 x + y))

\frac{d}{d x} (ln (5 x + 3))

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

\frac{d}{d x} (\frac{9 x}{e ^{x}})

f (x) = arccos (3 x)

What is the integral of (3x^39x^2-15)/(x^24x^4) ?
The integral of (3x^39x^2-15)/(x^24x^4) is 1/4 (27ln|x|+3/(x^5))+C