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

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

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

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

- \frac{1}{2 x ^{2}} - 2 x^{2} + x + C

Solution steps

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

Expand

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

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

\int \frac{1}{x ^{3}} d x = - \frac{1}{2 x ^{2}}

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

\int 1 d x = x

= - \frac{1}{2 x ^{2}} - 2 x^{2} + x

Add a constant to the solution

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

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