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Answer 1 · 2017-10-22T21:59:13

#(-1)/(3(3x+7))+C#

#int1/(3x+7)^2dx#

We can rewrite this:

#=int(3x+7)^-2dx#

Let's use the substitution #u=3x+7#. This implies that #du=3dx#.

#=1/3int(3x+7)^-2(3dx)#

#=1/3intu^-2du#

Now we can use the rule #intu^ndu=u^(n+1)/(n+1)+C#, where #n!=-1#.

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

#=(-1)/(3u)+C#

#=(-1)/(3(3x+7))+C#