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How do you integrate #int (3x^2)/sqrt(x^3+7)# using substitution?

Answer 1

Christopher Agresta

#2sqrt(x^3+7)+C#

#int(3x^2)/sqrt(x^3+7)dx#

Apply the substitution #x^3+7=u#. Thus, #3x^2dx=du#. Substituting in:

#=int(du)/sqrtu#

#=intu^(-1/2)du#

Using the rule #intu^ndu=u^(n+1)/(n+1)+C#, where #n!=-1#:

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

#=2sqrtu+C#

#=2sqrt(x^3+7)+C#

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Answer 2

Ezra Adams

To integrate (\int \frac{3x^2}{\sqrt{x^3 + 7}} , dx) using substitution, let (u = x^3 + 7). Then, (du = 3x^2 , dx). Substituting these into the integral yields:

(\int \frac{1}{\sqrt{u}} , du).

This simplifies to (\int u^{-1/2} , du), which integrates to (2u^{1/2} + C).

Finally, substituting back (u = x^3 + 7) gives the result:

(2\sqrt{x^3 + 7} + 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.