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

Answer 1

Please see the explanation.

Let #u = 5x + 4#, then #du = 5dx# or dx = #(1/5)du#
#int(5x + 4)^5dx = (1/5)intu^5du = u^6/30 + C#

Reverse the substitution:

#int(5x + 4)^5dx = (5x+4)^6/30 + C#
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Answer 2

Please see the explanation.

Let #u = 5x + 4#, then #du = 5dx# or dx = #(1/5)du#
#int(5x + 4)^5dx = (1/5)intu^5du = u^6/30 + C#

Reverse the substitution:

#int(5x + 4)^5dx = (5x+4)^6/30 + C#
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Answer 3

To integrate ( \int (5x+4)^5 ) using substitution, you can let ( u = 5x + 4 ). Then, find ( du ) by taking the derivative of ( u ) with respect to ( x ). Afterward, substitute ( u ) and ( du ) into the integral, and rewrite it in terms of ( u ). Finally, integrate the resulting expression with respect to ( u ), and then substitute back ( 5x + 4 ) for ( u ) to obtain the final answer.

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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.

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.

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.

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.

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