Question

Solve $\int \frac{dt}{{\left(1+t\right)}^2}$

Answer 1

$=\int \frac{dt}{{(1+t)}^2}$

Let $1+t=x$ and $dt=dx$

$=$$-\int \frac{-dx}{x^2}$

$=$ $\frac{-1}{x}$$+C$

$=\frac{1}{1+t}+C$