Question

Вычислите интеграл, пожалуйста​

Answer 1

$\int\limits {\frac{-7x^2+30x+4}{(x-4)\cdot(x+2)\cdot (x-5)}} \, dx \\ \\ \\ \frac{A}{x-4}+\frac{B}{x+2}+\frac{C}{x-5}=\frac{-7x^2+30x+4}{(x-4)\cdot(x+2)\cdot (x-5)} \\ \\ A(x+2)\cdot (x-5)+B\cdot (x-4)\cdot(x-5)+C\cdot (x-4)\cdot (x+2)=-7x^2+30x+4 \\ \\ A(x^2-3x-10)+B\cdot (x^2-9x+20)+C\cdot (x^2-2x-8)=-7x^2+30x+4 \\ \\ \begin{equation*} \begin{cases} A+B+C=-7 \\ -3A-9B-2C=30 \\ -10A+20B-8C=4 \end{cases}\end{equation*}$

$\begin{equation*} \begin{cases} C=-7-A-B \\ -3A-9B-2\cdot (-7-A-B)=30 \\ -10A+20B-8\cdot (-7-A-B)=4 \end{cases}\end{equation*} \\ \left \begin{equation*} \begin{cases} C=-7-A-B \\ -3A-9B+14+2A+2B=30 \\ -10A+20B+56+8A+8B=4 \end{cases}\end{equation*}\\ \left \begin{equation*} \begin{cases} C=-7-A-B \\ -A-7B=16 \\ -2A+28B=-52 \end{cases}\end{equation*}$

$\begin{equation*} \begin{cases} C=-7-A-B \\ A=-16-7B \\ -2\cdot (-16-7B)+28B=-52 \end{cases}\end{equation*} \\ \left \begin{equation*} \begin{cases} C=-7-A-B \\ A=-16-7B \\ 32+14B+28B=-52 \end{cases}\end{equation*}\\ \left \begin{equation*} \begin{cases} C=-7-A-B \\ A=-16-7B \\ 42B=-84 \end{cases}\end{equation*}\\ \left \begin{equation*} \begin{cases} C=-7-(-2)-(2)=-7+2+2=-3 \\ A=-16-7\cdot(-2)=-16+14=-2 \\ B=-2 \end{cases}\end{equation*}$

$\int\limits {\frac{-7x^2+30x+4}{(x-4)\cdot(x+2)\cdot (x-5)}} \, dx =\int {(\frac{-2}{x-4}+\frac{-2}{x+2}+\frac{-3}{x-5})} \, dx= \\ \\ =-\int {\frac{2}{x-4}} \. dx - \int {\frac{2}{x+2}} dx -\int{\frac{3}{x-5}} \, dx=-2\ln{|x-4|}-2\ln{|x+2|}-3\ln{|x-5|}+C =\\ \\ =-2\ln{|x^2-2x-8|}-3\ln{|x-5|}+C$