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

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

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

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

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

Solution steps

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

Simplify

x^{- 1} (2 x + x^{2}) : 2 + x

= \int 2 + x 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 2 d x + \int x d x

\int 2 d x = 2 x

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

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

Add a constant to the solution

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

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