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

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Answer 1 · 2018-04-16T03:00:34

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#