Find #intx(3x^2-1)^5# dx using implicit substitution. How?

Question

#intx(3x^2-1)^5#

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Answer 1 · 2017-05-04T11:53:21

The answer is #=1/36 (3x^2-1)^6+C#

We need

#intx^ndx=x^(n+1)/(n+1)+C(n!=-1)#

Let #u=3x^2-1#

Then,

#du=6xdx#

#xdx=1/6du#

Therefore,

#int(3x^2-1)^5xdx=1/6intu^5du#

#=1/36u^6#

#=1/36 (3x^2-1)^6+C#