Answer
$-\frac{\left(1+\frac{1}{t}\right)^4}{4}+C$
Work Step by Step
$\int\left(1+\frac{1}{t}\right)^3\left(\frac{1}{t^2}\right)dt$
Let $u=1+\frac{1}{t}$ then $du=-\frac{1}{t^2}dt$
so $dt=-t^2du$
The integral then becomes
$\int(u)^3\frac{1}{t^2}(-t^2)du=\int -u^3du=-\frac{u^4}{4}+C$
$=-\frac{\left(1+\frac{1}{t}\right)^4}{4}+C$
