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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{\frac{1}{2}}} d u

= \int \frac{1}{u ^{\frac{1}{2}}} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

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

Substitute back

u = 1 + x

= 2 (1 + x)^{\frac{1}{2}}

Add a constant to the solution

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

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