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How do you integrate #int (1+3t)t^2dt#?

1 Answer

Nov 28, 2016

The answer is #=t^3/3+(3t^4)/4+C#

Explanation:

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

So,

#int(1+3t)t^2dt=int (t^2+3t^3)dt#

#=intt^2dt+3intt^3dt#

#=t^3/3+3t^4/4+C#

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