What is the integral of (x^2+x+1)/(x^4) ?

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

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integral of (x^2+x+1)/(x^4)

\int \frac{x ^{2} + x + 1}{x ^{4}} d x

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

Solution steps

\int \frac{x ^{2} + x + 1}{x ^{4}} d x

Expand

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

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

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

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

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

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

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

Add a constant to the solution

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

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