How do you integrate #(5)/(1+3x)^3#?

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Answer 1 · 2015-08-23T12:18:39

The answer is #-5/(6(1+3x)^2)#

Rewrite as : #5/3int3/(1+3x)^3dx#

Now let's : #u = 1+3x#
then #du = 3#

So we have :

#5/3int1/u^3du = 5/3intu^(-3)du#

#=> 5/3[-1/2u^(-2)] = -5/6[1/u^2]#

Dont forget #u = 1+3x#

Finally : #-5/(6(1+3x)^2)#