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How do you integrate by substitution #int(x^2-9)^3(2x)dx#?

Answer 1

Ezra Adams

#(x^2-9)^4/4+c#

#y=x^2-9#

#dy=2xdx#

#int(x^2-9)^3(2x)dx=inty^3dy=y^4/4+c=(x^2-9)^4/4+c#

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Answer 2

Charles Anctil

To integrate the expression ∫(x^2-9)^3(2x)dx using the substitution method, let u = x^2 - 9. Then, du/dx = 2x.

Substitute u = x^2 - 9 into the integral, and replace dx with du/2x:

∫(x^2-9)^3(2x)dx = ∫u^3 du/2x.

Now, cancel out the common factor of 2x in the denominator:

∫u^3 du/2x = ∫u^3 du/(2(√u)).

Now integrate with respect to u:

∫u^3 du/(2(√u)) = (1/2)∫u^(3/2) du.

Now integrate u^(3/2) with respect to u:

(1/2)∫u^(3/2) du = (1/2) * (2/5) * u^(5/2) + C.

Substitute back u = x^2 - 9:

(1/2) * (2/5) * (x^2 - 9)^(5/2) + C = (1/5)(x^2 - 9)^(5/2) + C.

Therefore, the antiderivative of (x^2-9)^3(2x)dx is (1/5)(x^2 - 9)^(5/2) + 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.