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

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

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

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

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

Solution steps

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

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

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

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

\int \frac{1}{x} d x = 2 x

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

Add a constant to the solution

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

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