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 $n ne -1$

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