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How do you integrate #intx(4x+5)^3# using substitution?

Answer 1

Paisley Johnson

#=>1/80(4x + 5)^5 - 5/64(4x + 5)^4 + C#

Let #u = 4x + 5#. Then #du =4dx -> dx= (du)/4#. This also means that #x = (u - 5)/4#.

#=>int (u - 5)/4 u^3 (du)/4#

#=>1/16intu^4 - 5u^3du#

#=>1/16(1/5u^5 - 5/4u^4) + C#

#=>1/80u^5 - 5/64u^4 + C#

#=>1/80(4x + 5)^5 - 5/64(4x + 5)^4 + C#

Hopefully this helps!

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Answer 2

Emilia Aguilar

To integrate (\int (4x+5)^3 dx) using substitution, let (u = 4x + 5). Then, (du/dx = 4) or (dx = du/4). Substitute these into the integral to get (\int u^3 \cdot (1/4) du). Now, integrate (\int u^3 \cdot (1/4) du) to get ((1/4) \cdot (u^4/4) + C), where (C) is the constant of integration. Finally, substitute back (u = 4x + 5) to get the final answer of ((1/64)(4x+5)^4 + 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.