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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Simplify

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

Add a constant to the solution

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

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