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

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

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

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

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

Solution steps

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

Simplify

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

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

Expand

\frac{( 1 + x ^{2} ) ^{2}}{x ^{4}} : \frac{1}{x ^{4}} + \frac{2}{x ^{2}} + 1

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

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

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

\int 1 d x = x

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

Add a constant to the solution

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

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