有理函数的不定积分例题
example 0
0.First
\[\begin{aligned}
\int\frac{1}{\sin x+\cos x}dx=?
\\ \\
设:u=\tan\frac{x}{2}, \enspace x=2\arctan(u)
\\ \\
\sin x=\frac{2u}{1+u^{2}}, \enspace \cos x=\frac{1-u^{2}}{1+u^{2}}
\\ \\
\int\frac{1}{\frac{2u}{1+u^{2}}
+\frac{1-u^{2}}{1+u^{2}}}d\left(2\arctan u\right)
\\ \\
d(2\arctan u)=du\cdot(2\arctan u)'
\\ \\
[2\arctan(u)]^{\prime}=(2)^{\prime}\arctan(u)+2[\arctan(u)]'=
\frac{2}{1+u^{2}}
\\ \\
\int\frac{1+u^{2}}{2u+1-u^{2}}\cdot\frac{2}{1+u^{2}}du=2\int\frac{1}{1-u^{2}+2u}du
\end{aligned}
\]
0.Second
\[\begin{aligned}
1-u^{2}+2u\Rightarrow1+(u^{2}+2u)-1+1
\\ \\
1+(-u^{2}+2u-1)+1
\\ \\
\boxed {2+(u^{2}-2u+1)=2-(u-1)^{2} }
\\ \\
\boxed {[\sqrt{2}-(u-1)][\sqrt{2}+(u-1)]}
\\ \\
\Rightarrow
2\int\frac{1}{\left[\sqrt{2}-\left(u-1\right)\right]\left[\sqrt{2}+\left(u-1\right)\right]}du
\end{aligned}
\]
0.Thirdly
\[\begin{align}
设: u-1=j
\\ \\
\frac{A(\sqrt{2}+j)}{(\sqrt{2}-j)(\sqrt{2}+j)}+
\frac{B(\sqrt{2}-j)}{(\sqrt{2}-j)(\sqrt{2}+j)}
\\ \\
=\frac{A}{\sqrt{2}-j}+\frac{B}{\sqrt{2}+j}
\\ \\
=A(\sqrt{2}+j)+B(\sqrt{2}-j)=1
\\ \\
A\sqrt{2}+Aj+B\sqrt{2}-Bj=1
\\ \\
\sqrt{2}(A+B)+j(A-B)=1
\\ \\
\because j 为未知变量
\\ \\
\begin{cases}A-B=0\Rightarrow A=B
\\ \\
\sqrt{2}(A+B)=1 \end{cases}
\\ \\
2\sqrt{2}A=1\Rightarrow A=B=\frac{1}{2\sqrt{2}}
\\ \\
2\int\left(\frac{\frac{1}{2\sqrt{2}}}{\sqrt{2}-(u-1)}+\frac{\frac{1}{2\sqrt{2}}}{\sqrt{2}+(u-1)}\right)du
\\ \\
2\int \frac{1}{2\sqrt{2}} [\frac{1}{\sqrt{2}-(u-1)}+\frac{1}{\sqrt{2}+(u-1)}] du
\\ \\
\frac{1}{\sqrt{2}}\int\frac{1}{\sqrt{2}-(u-1)}du+\frac{1}{\sqrt{2}}\int\frac{1}{\sqrt{2}+(u-1)}du
\end{align}
\]
0.Fourth
\[\begin{align}
设:\sqrt{2}-(u-1)=h
\\ \\
\int\frac{1}{\sqrt{2}-(u+1)}du=\int\frac{1}{h}d\left[\sqrt{2}-(u-1)\right]
\\ \\
=\frac{1}{\sqrt{2}}\int\frac{1}{h}\cdot -1 dh=-\frac{1}{\sqrt{2}}\int\frac{1}{h}dh
\\ \\
=-\frac{1}{\sqrt{2}}\ln\left|h\right|+C
\\ \\
\boxed{ \frac{1}{\sqrt{2}}\ln\left|\sqrt{2}-\left(u-1\right)\right|+C }
\\ \\
\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{2}+(u-1)} du
\\ \\
\boxed{ \frac{1}{\sqrt{2}} \ln |\sqrt{2}+(u-1)|+C }
\end{align}
\]
0.Fifth
\[\begin{align}
-\frac{1}{\sqrt{2}} \ln |\sqrt{2}-(u-1)|+\frac{1}{\sqrt{2}} \ln |\sqrt{2}+(u-1)|
\\ \\
\frac{1}{\sqrt{2}} \ln |\sqrt{2}+(u-1)|-\frac{1}{\sqrt{2}} \ln |\sqrt{2}-(u-1)|
\\ \\
\frac{1}{\sqrt{2}} \left( \ln |\sqrt{2}+u-1|-\ln |\sqrt{2}-u+1| \right)+C
\\ \\
\frac{1}{\sqrt{2}}
\left(\ln \left|\tan \frac{x}{2}+\sqrt{2}-1\right|-\ln \left|\sqrt{2}+1-\tan \frac{x}{2}\right|\right)
+C
\\ \\
\frac{1}{\sqrt{2}}\ln\left|\frac{\tan\frac{x}{2}+\sqrt{2}-1}{\sqrt{2}+1-\tan\frac{x}{2}}\right|+C
\end{align}
\]
example1
first
\[\begin{align}
\int\frac{x^{2}+6}{x^{2}-9x+8}dx=?
\\ \\
对 x^{2}-9x+8 进行因式分解
\\ \\
据公式: x^{2}+(a+b)x+ab=(x+a)(x+b)
\\ \\
\begin{cases}
ab=8
\\ \\
a+b=-9
\end{cases}
\\ \\
a=-1,b=-8
\\ \\
x^{2}-9y+8\Rightarrow (x-1)(x-8)
\\ \\
\frac{x+b}{(x-1)(x-8)}\Rightarrow\frac{x}{(x-1)(x-8)}+\frac{6}{(x-1)(x-8)}
\\ \\
转化为:\frac{A(x-8)}{(x-1)(x-8)}+\frac{B(x-1)}{(x-1)(x-8)}=\frac{A}{(x-1)}+\frac{B}{(x-8)}
\end{align}
\]
second
\[
\begin{aligned}
x+6=A(x-8)+B(x-1)
\\ \\
=Ax-8A+Bx-B
\\ \\
Ax+Bx-8A-B
\\ \\
x+Bx=x(A+B)=x\Rightarrow A+B=1
\\ \\
-8A-B=6
\\ \\
\because B=1-A
\\ \\
-8A-(1-A)=6,A=-1,B=2
\\ \\
\Rightarrow\frac{-1}{x-1}+\frac{2}{x-8}
\end{aligned}
\]
thirdly
\[\begin{aligned}
\int\frac{-1}{x-1}dx+\int\frac{2}{x-8}dx
\\ \\
\int\frac{-1}{x-1}dx
\\
设:u=x-1
\\ \\
\Rightarrow-1\int\frac{1}{u}d(x-1)
\\
d(x-1)=du\cdot(x-1)^{\prime}=du
\\ \\
-\int\frac{1}{u}du=-\ln|u|+C
\\ \\
\int\frac{2}{x-8}dx
\\
设:V=x-8
\\ \\
\Rightarrow2\int\frac{1}{V}d(x-8)
\\
d(x-8)=dV\cdot(x-8)^{\prime}=dV
\\ \\
2\int\frac{1}{V}dV=2\ln \vert V\vert+C
\\ \\
-\ln\left|u\right|+2\ln\left|V\right|+C
\\ \\
-\ln\left|x-1\right|+2\ln\left|x-8\right|+C
\end{aligned}
\]
example2
first
\[\begin{align}
\int \frac{1}{(2+5 x)\left(x^{2}+7\right)} d x =?
\\ \\
\frac{A\left(x^{2}+7\right)}{(2+5 x)\left(x^{2}+7\right)}+\frac{M x+N}{(2+5 x)\left(x^{2}+7\right)} \Rightarrow \frac{A}{2+5 x}+\frac{M x+N}{x^{2}+7}
\\ \\
A\left(x^{2}+7\right)+(Mx+N)(2+5 x)=1
\\ \\
A x^{2}+7 A+2 Mx+5 M x^{2}+2 N+5 Nx=1
\\ \\
x^{2}(A+5 M)+x(2 M+5 N)+7 A+2 N=1
\\ \\
\left\{\begin{array}{l}
A+5 M=0
\\ \\
2 M+5 N=0
\\ \\
7 A+2 N=1
\end{array}\right.
\\ \\
A=-5 M, \quad M=-\frac{5}{2}N, \quad A=\frac{1-2 N}{7}
\\ \\
\text { 据: } \quad A+5M=0 \Rightarrow \frac{1-2 N}{7}+5\left(-\frac{5 N}{2}\right)=0
\\ \\
\frac{2-4 N}{14}-\frac{175 N}{14}=0
\\ \\
N=\frac{2}{179}, \quad M=-\frac{5}{179}, \quad A=\frac{25}{179}
\\ \\
\Rightarrow \frac{\frac{25}{179}}{2+5 x}+\frac{-\frac{5}{179} x+\frac{2}{179}}{x^{2}+7}
\end{align}
\]
second
\[\begin{align}
\int\frac{25}{179}dx,
\\
设:u=2+5x
\\ \\
\frac{25}{179}\int\frac{1}{u}d(2+5x)=\frac{25}{179}\int\frac{1}{u}\cdot5du
\\ \\
\frac{125}{179}\int\frac{1}{u}du=\frac{125}{179}\ln|u|+C
\\ \\
\frac{125}{179}\ln\vert2+5x\vert+C
\\ \\
\\ \\
\int\frac{-5}{\frac{179}{x^{2}+7}}dx
\\ \\
设:v=x^{2}+7
\\ \\
d(v)=(v)^{\prime}\cdot dv=(x^{2}+7)^{\prime}\cdot dv=2xdv
\\ \\
-\frac{5}{179}\int\frac{x}{v}2xdv=-\frac{10}{179}\int\frac{1}{v}dv
\\ \\
=-\frac{10}{179}\ln\left|v\right|+C=\frac{-10}{179}\ln\left|x^{2}+7\right|+C
\end{align}
\]
thirdly
\[\begin{align}
\int \frac{\frac{2}{179}}{x^{2}+7} d x
\\ \\
\frac{2}{179} \int \frac{1}{\left(\frac{x^{2}}{7}+1\right) 7} d x \Rightarrow \frac{2}{179} \int \frac{1}{7} \cdot \frac{1}{1+\left(\frac{x}{\sqrt{7}}\right)^{2}} d x
\\ \\
\text { 设: } t=\frac{x}{\sqrt{7}}
\\ \\
\frac{2}{179} \int \frac{1}{7} \cdot \frac{1}{1+t^{2}} d\left(\frac{x}{\sqrt{7}}\right)
\\ \\
d\left(\frac{x}{\sqrt{7}}\right)=d x\left(\frac{x}{\sqrt{7}}\right)^{\prime}=\frac{1}{\sqrt{7}} d x
\\ \\
d t=\frac{1}{\sqrt{7}} d x, \enspace d x=\sqrt{7} d t
\\ \\
\frac{2}{179} \int \frac{1}{7} \cdot \sqrt{7} \cdot \frac{1}{1+t^{2}} d t=\frac{2}{179 \sqrt{7}} \int \frac{1}{1+t^{2}} d t
\\ \\
据积分表中的公式: \int \frac{1}{1+x^{2}}dx=\arctan(x)+C
\\ \\
\Rightarrow \frac{2}{179 \sqrt{7}} \arctan (t)+C
\\ \\
=\frac{2}{179 \sqrt{7}} \arctan \left(\frac{x}{\sqrt{7}}\right)+C
\\ \\
\\ \\
\Rightarrow \frac{125}{179} \ln |2+5 x|-\frac{10}{179} \ln \left|x^{2}+7\right|+\frac{2}{179 \sqrt{7}} \arctan \left(\frac{x}{\sqrt{7}}\right) + C
\end{align}
\]
