How do you integrate by substitution #int (t^3/3+1/(4t^2))dt#?

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Answer 1 · 2018-06-19T18:45:45

#=t^4/12-1/(4t)+c#

you don't do this by substitution just use the power rule

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

#int(t^3/3+1/(4t^2))dt#

#=int(t^3/3+1/4t^(-2))dt#

#=1/3(t^4)/4+(1/4)t^(-1)/(-1)+c#

#=t^4/12-1/(4t)+c#

Answer 2 · 2018-06-19T18:48:20

#t^4/12-1/(4t)+C#

Writing
#int( t^3/3+1/4t^(-2))dt#
we get

#t^4/12-1/(4t)+C#
we are using that

#int x^ndx=x^(n+1)/(n+1)+C# if #n ne -1#