Задача №1644
Условие
Найти интеграл \(\int\frac{dx}{\left(1-2^x\right)^4}\).
Решение
\[
\int\frac{dx}{\left(1-2^x\right)^4}
=\int\frac{2^xdx}{2^x\left(1-2^x\right)^4}
=\left[\begin{aligned}
& t=1-2^x;\;2^x=1-t.\\
& dt=-2^x\ln{2}dx;\; 2^xdx=-\frac{dt}{\ln{2}}.
\end{aligned}\right]
=\frac{1}{\ln{2}}\int\frac{dt}{(t-1)t^4}.
\]
\[
\frac{1}{(t-1)t^4}
=\frac{A_0}{t-1}+\frac{A_1}{t}+\frac{A_2}{t^2}+\frac{A_3}{t^3}+\frac{A_4}{t^4}
=\frac{A_0t^4+A_1(t-1)t^3+A_2(t-1)t^2+A_3(t-1)t+A_4(t-1)}{(t-1)t^4}
\]
\[
\begin{aligned}
& 1=A_0t^4+A_1(t-1)t^3+A_2(t-1)t^2+A_3(t-1)t+A_4(t-1)\\
& t=1;\;A_0=1.
\end{aligned}
\]
Подставляя \(A_0=1\), будем иметь:
\[
1-t^4=A_1(t-1)t^3+A_2(t-1)t^2+A_3(t-1)t+A_4(t-1);\\
-t^3-t^2-t-1=A_1t^3+A_2t^2+A_3t+A_4;\\
A_1=-1;\;A_2=-1;\;A_3=-1;\;A_4=-1.
\]
\[
\frac{1}{(t-1)t^4}
=\frac{1}{t-1}-\frac{1}{t}-\frac{1}{t^2}-\frac{1}{t^3}-\frac{1}{t^4}
\]
\[
\frac{1}{\ln{2}}\int\frac{dt}{(t-1)t^4}
=\frac{1}{\ln{2}}\int\left(\frac{1}{t-1}-\frac{1}{t}-\frac{1}{t^2}-\frac{1}{t^3}-\frac{1}{t^4}\right)dt=\\
=\frac{1}{\ln{2}}\cdot\left(\ln|t-1|-\ln|t|+\frac{1}{t}+\frac{1}{2t^2}+\frac{1}{3t^3}\right)+C=\\
=x-\log_2\left|1-2^x\right|+\frac{1}{\ln{2}}\cdot\left(\frac{1}{1-2^x}+\frac{1}{2\left(1-2^x\right)^2}+\frac{1}{3\left(1-2^x\right)^3}\right)+C
\]
Ответ:
\(x-\log_2\left|1-2^x\right|+\frac{1}{\ln{2}}\cdot\left(\frac{1}{1-2^x}+\frac{1}{2\left(1-2^x\right)^2}+\frac{1}{3\left(1-2^x\right)^3}\right)+C\)