Задача №1935

\[ \int\frac{dx}{\ch^5{x}} =\left[\begin{aligned} & u=\frac{1}{\ch^3{x}};\;du=-\frac{3\sh{x}}{\ch^4{x}}dx;\\ & dv=\frac{dx}{\ch^2{x}};\;v=\th{x}. \end{aligned}\right] =\frac{\th{x}}{\ch^3{x}}+3\int\frac{\th{x}\cdot\sh{x}}{\ch^4{x}}dx=\\ =\frac{\th{x}}{\ch^3{x}}+3\int\frac{\sh^2{x}}{\ch^5{x}}dx =\frac{\th{x}}{\ch^3{x}}+3\int\frac{\ch^2{x}-1}{\ch^5{x}}dx =\frac{\th{x}}{\ch^3{x}}+3\int\frac{dx}{\ch^3{x}}-3\int\frac{dx}{\ch^5{x}}. \]

\[ \int\frac{dx}{\ch^5{x}} =\frac{\th{x}}{4\ch^3{x}}+\frac{3}{4}\cdot\int\frac{dx}{\ch^3{x}} \tag{1}\label{1} \]

\[ \int\frac{dx}{\ch^3{x}} =\left[\begin{aligned} & u=\frac{1}{\ch{x}};\;du=-\frac{3\sh{x}}{\ch^2{x}}dx;\\ & dv=\frac{dx}{\ch^2{x}};\;v=\th{x}. \end{aligned}\right] =\frac{\th{x}}{\ch{x}}+\int\frac{\th{x}\cdot\sh{x}}{\ch^2{x}}dx=\\ =\frac{\th{x}}{\ch{x}}+\int\frac{\sh^2{x}}{\ch^3{x}}dx =\frac{\th{x}}{\ch{x}}+\int\frac{\ch^2{x}-1}{\ch^3{x}}dx =\frac{\th{x}}{\ch{x}}+\int\frac{dx}{\ch{x}}-\int\frac{dx}{\ch^3{x}}. \]

\[ \int\frac{dx}{\ch^3{x}} =\frac{\th{x}}{2\ch{x}}+\frac{1}{2}\int\frac{dx}{\ch{x}} =\frac{\th{x}}{2\ch{x}}+\int\frac{dx}{e^x+e^{-x}} =\frac{\th{x}}{2\ch{x}}+\int\frac{e^xdx}{e^{2x}+1} =\frac{\th{x}}{2\ch{x}}+\arctg{e^x}+C. \]

\[ \int\frac{dx}{\ch^5{x}} =\frac{\th{x}}{4\ch^3{x}}+\frac{3\th{x}}{8\ch{x}}+\frac{3\arctg{e^x}}{4}+C \]