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

Answer 1

Aurora Alvarado

#(2/3)^2(x^(3/2)-1)^(3/2) + C #

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

#sqrtxsqrt(xsqrtx+1) = x^(1/2)(x^(3/2)-1)^(1/2)=(2/3)^2d/dx(x^(3/2)-1)^(3/2)#

#int sqrtx(sqrt(xsqrtx+1))dx = (2/3)^2(x^(3/2)-1)^(3/2) + C #

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Answer 2

Aurora Alvarado

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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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.