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How do you integrate #int(5x^4+5)^(2/3)*20x^3 dx# using substitution?

Answer 1

Emily Ammerman

Let #u = 5x^4 + 5#, then #(du)/dx = 20x^3 -> du = 20x^3dx -> dx = (du)/(20x^3)#

#=int(u^(2/3)) xx 20x^3 xx (du)/(20x^3)#

#=int(u^(2/3)) xx cancel(20x^3) xx (du)/cancel(20x^3)#

#=(u^(2/3 + 1))/(2/3 + 1) + C#

#= (u^(5/3))/(5/3) + C#

#= 3/5u^(5/3) + C#

#=3/5(5x^4 + 5)^(5/3) + C#

Hopefully this helps!

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Answer 2

William Abdalla

To integrate ( \int (5x^4+5)^{\frac{2}{3}} \cdot 20x^3 , dx ) using substitution, let ( u = 5x^4 + 5 ). Then, ( du = 20x^3 dx ). Substituting, the integral becomes ( \int u^{\frac{2}{3}} , du ). Now, integrate ( u^{\frac{2}{3}} ) with respect to ( u ). Finally, substitute back ( u ) with ( 5x^4 + 5 ) to get the final result.

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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.