What is the integral of (x^{-1}-x^{-2}+x^{-3})/(x^2) ?

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

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integral of (x^{-1}-x^{-2}+x^{-3})/(x^2)

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

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

Solution steps

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

Expand

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

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

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

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

Simplify

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

Add a constant to the solution

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

\int \frac{( x ^{4} )}{x - 1} d x

\int cos^{4} (9 x) d x

(ln (x))^{3}

\int tan^{3} (4 x) d x

8 - 3 x, 6 - x, 3

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