Question

Find teh indefinite integral of
integral of [(1+ 1/t)^3 (1/t^2) dt

Answer 1

If I am reading your equation correctly:
\[\int\left(1+\frac{1}{t}\right)^3\frac{1}{t^2}dt =
\int\left(1+\frac{3}{t}+\frac{3}{t^2}+\frac{1}{t^3}\right)\frac{1}{t^2}dt\]
\[=\int\left(t^{-2}+3t^{-3}+3t^{-4}+t^{-5}\right)dt\]
Then you just need to use the exponent integration rules.

Answer 2

um so u mean like it becomes
(t^-1/ -1) + (3t^-2/-2) + (3t^-3/-3) + (t^-4/-4)

Answer 3

Yep. Of course, you can check your work by solving this with a u-substitution \[u = \frac{1}{t}\] and \[du = -\frac{1}{t^2}dt\]
so....
\[\int\left(1+t^{-1}\right)^3t^{-2}dt = -\int\left(1+u\right)^3du\]