What is the integral of (-8x^4+3x^2+9)/(3x^3) ?

The integral of (-8x^4+3x^2+9)/(3x^3) is -4/3 x^2+ln|x|-3/(2x^2)+C

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integral of (-8x^4+3x^2+9)/(3x^3)

\int \frac{- 8 x ^{4} + 3 x ^{2} + 9}{3 x ^{3}} d x

- \frac{4}{3} x^{2} + ln ∣ x ∣ - \frac{3}{2 x ^{2}} + C

Solution steps

\int \frac{- 8 x ^{4} + 3 x ^{2} + 9}{3 x ^{3}} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{- 8 x ^{4} + 3 x ^{2} + 9}{x ^{3}} d x

Expand

\frac{- 8 x ^{4} + 3 x ^{2} + 9}{x ^{3}} : - 8 x + \frac{3}{x} + \frac{9}{x ^{3}}

= \frac{1}{3} \cdot \int - 8 x + \frac{3}{x} + \frac{9}{x ^{3}} 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}{3} (- \int 8 x d x + \int \frac{3}{x} d x + \int \frac{9}{x ^{3}} d x)

\int 8 x d x = 4 x^{2}

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

= \frac{1}{3} (- 4 x^{2} + 3 ln ∣ x ∣ - \frac{9}{2 x ^{2}})

Simplify

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

= - \frac{4}{3} x^{2} + ln ∣ x ∣ - \frac{3}{2 x ^{2}}

Add a constant to the solution

= - \frac{4}{3} x^{2} + ln ∣ x ∣ - \frac{3}{2 x ^{2}} + C

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What is the integral of (-8x^4+3x^2+9)/(3x^3) ?
The integral of (-8x^4+3x^2+9)/(3x^3) is -4/3 x^2+ln|x|-3/(2x^2)+C