What is the integral of (2x^8-7x^4)/(3x^9) ?

The integral of (2x^8-7x^4)/(3x^9) is 1/3 (2ln|x|+7/(4x^4))+C

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integral of (2x^8-7x^4)/(3x^9)

\int \frac{2 x ^{8} - 7 x ^{4}}{3 x ^{9}} d x

\frac{1}{3} (2 ln ∣ x ∣ + \frac{7}{4 x ^{4}}) + C

Solution steps

\int \frac{2 x ^{8} - 7 x ^{4}}{3 x ^{9}} 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{2 x ^{8} - 7 x ^{4}}{x ^{9}} d x

Simplify

\frac{2 x ^{8} - 7 x ^{4}}{x ^{9}} : \frac{2 x ^{4} - 7}{x ^{5}}

= \frac{1}{3} \cdot \int \frac{2 x ^{4} - 7}{x ^{5}} d x

Expand

\frac{2 x ^{4} - 7}{x ^{5}} : \frac{2}{x} - \frac{7}{x ^{5}}

= \frac{1}{3} \cdot \int \frac{2}{x} - \frac{7}{x ^{5}} 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 \frac{2}{x} d x - \int \frac{7}{x ^{5}} d x)

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int \frac{7}{x ^{5}} d x = - \frac{7}{4 x ^{4}}

= \frac{1}{3} (2 ln ∣ x ∣ - (- \frac{7}{4 x ^{4}}))

Simplify

= \frac{1}{3} (2 ln ∣ x ∣ + \frac{7}{4 x ^{4}})

Add a constant to the solution

= \frac{1}{3} (2 ln ∣ x ∣ + \frac{7}{4 x ^{4}}) + C

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What is the integral of (2x^8-7x^4)/(3x^9) ?
The integral of (2x^8-7x^4)/(3x^9) is 1/3 (2ln|x|+7/(4x^4))+C