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

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

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

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

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

Solution steps

\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

\int 4 x e^{- 5 x} d x

x \to a lim (\frac{1}{x})

\int sin (5 t) d t

x \to - 5 lim (\frac{1}{x + 5})

\frac{d}{d x} (\frac{3}{5} x^{15})

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