# How do you integrate #int sqrtxsqrt(xsqrtx+1)# using substitution?

This integral is a power integral. The square root operator helps to conceal de true problem.

So

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To integrate ( \int \sqrt{x\sqrt{x+1}} ) using substitution:

- Let ( u = \sqrt{x+1} ).
- Then, ( x = u^2 - 1 ).
- Find ( dx ) in terms of ( du ): [ dx = 2u , du ]
- Substitute ( u = \sqrt{x+1} ) and ( dx = 2u , du ) into the integral: [ \int \sqrt{x\sqrt{x+1}} , dx = \int u \cdot u \cdot 2u , du = 2\int u^3 , du ]
- Integrate ( u^3 ) with respect to ( u ): [ 2\int u^3 , du = 2\left(\frac{u^4}{4}\right) + C = \frac{1}{2}u^4 + C ]
- Substitute back ( u = \sqrt{x+1} ): [ \frac{1}{2}u^4 + C = \frac{1}{2}(\sqrt{x+1})^4 + C = \frac{1}{2}(x+1)^2 + C ]

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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.

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.

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.

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