What is the integral of (x^3-8x+1)/(8x^4) ?

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

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integral of (x^3-8x+1)/(8x^4)

\int \frac{x ^{3} - 8 x + 1}{8 x ^{4}} d x

\frac{1}{8} (ln ∣ x ∣ + \frac{4}{x ^{2}} - \frac{1}{3 x ^{3}}) + C

Solution steps

\int \frac{x ^{3} - 8 x + 1}{8 x ^{4}} d x

Take the constant out:

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

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

Expand

\frac{x ^{3} - 8 x + 1}{x ^{4}} : \frac{1}{x} - \frac{8}{x ^{3}} + \frac{1}{x ^{4}}

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

\int \frac{1}{x ^{4}} d x = - \frac{1}{3 x ^{3}}

= \frac{1}{8} (ln ∣ x ∣ - (- \frac{4}{x ^{2}}) - \frac{1}{3 x ^{3}})

Simplify

= \frac{1}{8} (ln ∣ x ∣ + \frac{4}{x ^{2}} - \frac{1}{3 x ^{3}})

Add a constant to the solution

= \frac{1}{8} (ln ∣ x ∣ + \frac{4}{x ^{2}} - \frac{1}{3 x ^{3}}) + C

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