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# 有理函数的积分，并非只有部分分式法

$\int\frac{1}{x^7-x}dx,\qquad \int\frac{x^2-1}{x^4+1}$

$\int\frac{1}{x^7-x}dx$

$\frac{1}{x^7-x} =\frac{1}{x(x^3-1)(x^3+1)}=\frac{1}{x(x-1)(x^2+x+1)(x+1)(x^2-x+1)}$

$\frac{A_1}{x}, \frac{A_2}{x-1},\frac{A_3}{x+1},\frac{B_1x+C_1}{x^2+x+1},\frac{B_2x+C_2}{x^2-x+1}$

$\int\frac{1}{x^7-x}dx=\int\frac{1}{x^7(1-\frac{1}{x^6})}dx$

\begin{align*} \int\frac{1}{x^7(1-\frac{1}{x^6})}dx &=\frac{1}{6}\int\frac{6}{x^7}\frac{1}{1-\frac{1}{x^6}}dx\\ &= \int\ \frac{1}{6}\frac{1}{u}du\\ &=\frac{1}{6}\ln|u|+c \\ &=\frac{1}{6}\ln|1-\frac{1}{x^6}|+c\end{align*}

$\int\frac{x^2-1}{x^4-1}dx$

$\int\frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx$

$\int\frac{ 1-\frac{1}{x^2} }{(x+\frac{1}{x})^2-2}dx$

$\int\frac{du}{u^2-2}$

\begin{align*}\int\frac{du}{u^2-2} &= \frac{1}{2\sqrt2} \int\frac{du}{u-\sqrt2}- \frac{1}{2\sqrt2} \int\frac{du}{u+\sqrt2}\\ &= \frac{1}{2\sqrt2}( \ln|u-\sqrt2|+\ln|u+\sqrt2|)+C\\ &= \frac{1}{2\sqrt2} \ln\left|\frac{u-\sqrt2}{u+\sqrt2} \right|+C \end{align*}

$\int\frac{x^2-1}{x^4+1}dx= \frac{1}{2\sqrt2} \ln\left|\frac{x+\frac{1}{x}-\sqrt2}{x+\frac{1}{x}+\sqrt2} \right|+C$

\begin{align*}\frac{x^2-1}{x^4+1}&=\frac{x^2-1}{(x^4+2x^2+1)-2x^2}\\ &=\frac{x^2-1}{(x^2+1)^2-2x^2}\\ &= \frac{x^2-1}{(x^2+1-\sqrt{2}x)(x^2+1-\sqrt{2}x)}\\ &= \frac{A_1x+B_1}{x^2+1-\sqrt{2}x}+\frac{A_2x+B_2}{x^2+1-\sqrt{2}x} \end{align*}