What is the integral of (x^3-2x^2+x+3)/(x^5) ?

The integral of (x^3-2x^2+x+3)/(x^5) is -1/x+1/(x^2)-1/(3x^3)-3/(4x^4)+C

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integral of (x^3-2x^2+x+3)/(x^5)

\int \frac{x ^{3} - 2 x ^{2} + x + 3}{x ^{5}} d x

- \frac{1}{x} + \frac{1}{x ^{2}} - \frac{1}{3 x ^{3}} - \frac{3}{4 x ^{4}} + C

Solution steps

\int \frac{x ^{3} - 2 x ^{2} + x + 3}{x ^{5}} d x

Expand

\frac{x ^{3} - 2 x ^{2} + x + 3}{x ^{5}} : \frac{1}{x ^{2}} - \frac{2}{x ^{3}} + \frac{1}{x ^{4}} + \frac{3}{x ^{5}}

= \int \frac{1}{x ^{2}} - \frac{2}{x ^{3}} + \frac{1}{x ^{4}} + \frac{3}{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

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

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

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

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

\int \frac{3}{x ^{5}} d x = - \frac{3}{4 x ^{4}}

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

Simplify

= - \frac{1}{x} + \frac{1}{x ^{2}} - \frac{1}{3 x ^{3}} - \frac{3}{4 x ^{4}}

Add a constant to the solution

= - \frac{1}{x} + \frac{1}{x ^{2}} - \frac{1}{3 x ^{3}} - \frac{3}{4 x ^{4}} + C

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What is the integral of (x^3-2x^2+x+3)/(x^5) ?
The integral of (x^3-2x^2+x+3)/(x^5) is -1/x+1/(x^2)-1/(3x^3)-3/(4x^4)+C