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

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

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

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

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

Solution steps

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

Simplify

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

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

Expand

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

Add a constant to the solution

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

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