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

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

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

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

- \frac{3}{2 x ^{2}} - ln ∣ x ∣ - \frac{1}{3 x ^{3}} + C

Solution steps

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

\frac{3 x - x ^{3} + 1}{x ^{4}} = (\frac{3 x}{x ^{4}} - \frac{x ^{3}}{x ^{4}} + \frac{1}{x ^{4}})

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

Simplify

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

= \int \frac{3}{x ^{3}} - \frac{1}{x} + \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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= - \frac{3}{2 x ^{2}} - ln ∣ x ∣ - \frac{1}{3 x ^{3}}

Add a constant to the solution

= - \frac{3}{2 x ^{2}} - ln ∣ x ∣ - \frac{1}{3 x ^{3}} + C

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