How do you integrate #y = (5x + 3)^6#?

Answer 1

#=1/35*(5x+3)^7+C#

So you've got...

#int(5x+3)^6dx#

Now, make:

#u=5x+3#

Which means that:

#(du)/(dx)=5#

And as a result:

#(dx)/(du)=1/5#
#dx=1/5du#

Now, transform:

#int(5x+3)^6dx# into #1/5intu^6du#

Integrate:

#1/5intu^6du#
#=1/5*u^7/7+C#
#=1/35*u^7+C#
#=1/35*(5x+3)^7+C#
#=int(5x+3)^6dx#
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Answer 2

To integrate the function ( y = (5x + 3)^6 ), you can use the substitution method. Let ( u = 5x + 3 ), then ( du/dx = 5 ) or ( du = 5dx ). Rearrange to solve for ( dx ), giving ( dx = du/5 ). Now substitute ( u ) and ( du/5 ) into the integral. The integral becomes ( \int u^6 \cdot (1/5) \cdot du ). Now integrate ( u^6 ) with respect to ( u ) to get ( (1/7)u^7 ), then multiply by ( 1/5 ) to get the final answer of ( (1/35)(5x + 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.

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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