What is the integral of 10x^{-1/2}+1 ?

The integral of 10x^{-1/2}+1 is 20x^{1/2}+x+C

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integral of 10x^{-1/2}+1

\int 10 x^{- \frac{1}{2}} + 1 d x

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

Solution steps

\int 10 x^{- \frac{1}{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 10 x^{- \frac{1}{2}} d x + \int 1 d x

\int 10 x^{- \frac{1}{2}} d x = 20 x^{\frac{1}{2}}

\int 1 d x = x

= 20 x^{\frac{1}{2}} + x

Add a constant to the solution

= 20 x^{\frac{1}{2}} + x + C

2 \frac{1}{x}, at x = 3

\int \frac{9 x ^{2} - 1}{x} d x

\int (1 - 2 e^{t a n (x)}) sec^{2} (x) d x

x \to 4 - lim (\frac{x + 2}{x - 4})

x \to 0 lim (\frac{lo g _{e} ( 1 + 3 x ^{2} )}{x ( e ^{x} - 1 )})

What is the integral of 10x^{-1/2}+1 ?
The integral of 10x^{-1/2}+1 is 20x^{1/2}+x+C