What is the integral of y= 1/(x^{1/2)} ?

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

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integral of y= 1/(x^{1/2)}

integral

y = \frac{1}{x ^{\frac{1}{2}}}

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

Solution steps

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

Apply the Power Rule

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

Add a constant to the solution

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

in t e g r a l 3 x + 1

in t e g r a l x^{8} + 9

in t e g r a l (9 x + 36)^{- 1}

in t e g r a l \frac{y ^{3}}{x ^{4}}

in t e g r a l \frac{1}{x ( 1 - x )}

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