How do you integrate by substitution $int t^2(t-2/t)dt$ ?

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Answer 1 · 2017-05-28T03:57:15

$((t^2-2)^2)/4+C$

$intt^2(t-2/t)dt$
$intt(t^2-2)dt$

Substitute $u=t^2-2$ , $(du)/(dx) =2t$

$1/2int2t(t^2-2)dt$
$1/2int(u)du$
$1/2((u^2)/2)$
$(u^2)/4$

Undo substitution:
$((t^2-2)^2)/4+C$

How do you integrate by substitution int t^2(t-2/t)dtint t^2(t-2/t)dt?