How do you integrate $int (1+1/t)^3*1/t^2$ using substitution?

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Answer 1 · 2016-07-21T23:20:33

$int (1+1/t)^3 * 1/t^2 dt =-(t+1)^4/(4t) + C$

Let $u = 1 + 1/t implies du = -1/t^2dt$

$int (1+1/t)^3 * 1/t^2 dt = - int u^3du$

$-1/4u^4+C = -1/4(1+1/t)^4 +C$

Tidying up a little gives:

$=-(t+1)^4/(4t) + C$

How do you integrate int (1+1/t)^3*1/t^2int (1+1/t)^3*1/t^2 using substitution?