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

1 Answer

May 28, 2017

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

Explanation:

#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#

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