Пр.1

$$ \int \frac{dt}{t} = ln |t|+C \\

\int \frac{dt}{1+t} = \int \frac{d(1+t)}{1+t} = ln|1+t|+C \\

\int \frac{dt}{1-t} = \int \frac{d(1-t)}{1-t} = ln|1-t|+C \\

\int\frac{dt}{t(t+1)} = \int \frac{Adt}{t} + \int\frac{Bdt}{t+1} \\

\frac{1}{t(t+1)} = \frac{A}{t} + \frac{B}{t+1} \\

1 = A(t+1) + Bt \Leftrightarrow (A+B)t + A-1 = 0 \\

\left\{\begin{matrix} A+B = 0 \\ A-1 = 0 \end{matrix}\right.

\left\{\begin{matrix} B = -A = -1 \\ A=1 \end{matrix}\right. \\

\int \frac{dt}{t(t+1)} = \int \frac{dt}{t} - \int \frac{dt}{t+1} = \ln |t| - ln|t+1|+C\\

$$