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

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

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

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

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

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

= \int (x + 1)^{- \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 = x + 1

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

Add a constant to the solution

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

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