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How do you use substitution to integrate #4 / ((3x + 6)^2) dx#?

Answer 1

Emilia Aguilar

use the u substitution

#u=3x+6#

#du=3dx#

to get #4# multiply it by #4/3#

#4/3*du=4/cancel(3)*cancel(3)dx#

#4/3du=4dx#

bring the integral into the u world

#int(4/3)/u^2#

bring the constant into the front

#4/3int1/u^2#

use the index law

#1/y^z=y^-z#

#4/3intu^-2#

use the power rule for integration and integrate

#intx^n=x^(n+1)/(n+1)#

#=4/3intu^-2#

#4/3(u^(-2+1)/(-2+1))+C#

#4/3(u^(-1)/(-1))+C#

bring it into the #x# world

#4/3(-(3x+6)^-1)+C#

#4/3(-1/(3x+6))+C#

finally multiply it

#-4/(9x+18)+C#

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Answer 2

Aurora Alvarado

To integrate ( \frac{4}{(3x + 6)^2} ) using substitution, let:

[ u = 3x + 6 ]

Then, find ( du ) in terms of ( dx ):

[ du = 3dx ]

Rearrange to solve for ( dx ):

[ dx = \frac{du}{3} ]

Now substitute ( u ) and ( du ) into the integral:

[ \int \frac{4}{u^2} \cdot \frac{du}{3} ]

[ = \frac{4}{3} \int \frac{1}{u^2} \ du ]

Now integrate with respect to ( u ):

[ = \frac{4}{3} \left( -\frac{1}{u} \right) + C ]

Substitute back ( u = 3x + 6 ) to get the final answer:

[ = -\frac{4}{3(3x + 6)} + 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.