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How do you find the indefinite integral of #int 1/sqrt(x+1)#?

Answer 1

Ezra Adams

Let #u = x + 1#. Then #du = dx#.

#=int 1/sqrt(u)du#

#= int u^(-1/2) du#

#=2u^(1/2) + C#

#=2sqrt(x + 1) + C#

Hopefully this helps!

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Answer 2

Cameron Alexander

To find the indefinite integral of ( \int \frac{1}{\sqrt{x+1}} ), you can use the method of u-substitution. Let ( u = x + 1 ). Then, ( du = dx ).

Now, rewrite the integral in terms of ( u ): [ \int \frac{1}{\sqrt{x+1}} dx = \int \frac{1}{\sqrt{u}} du ]

This integral can be rewritten as: [ \int u^{-\frac{1}{2}} du ]

Now, integrate ( u^{-\frac{1}{2}} ) with respect to ( u ): [ \int u^{-\frac{1}{2}} du = 2u^{\frac{1}{2}} + C ]

Finally, substitute ( u = x + 1 ) back into the equation: [ 2(x+1)^{\frac{1}{2}} + C ]

So, the indefinite integral of ( \int \frac{1}{\sqrt{x+1}} ) is ( 2(x+1)^{\frac{1}{2}} + C ), where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.