前置知识：第二类换元法
计算∫1x8(x2+1)dx\int\dfrac{1}{x^8(x^2+1)}dx∫x8(x2+1)1dx
解：
\qquad令x=1tx=\dfrac 1tx=t1，t=1xt=\dfrac 1xt=x1，dx=−1t2dtdx=-\dfrac{1}{t^2}dtdx=−t21dt
\qquad原式=∫11t8⋅(1t2+1)⋅(−1t2)dt=\int\dfrac{1}{\frac{1}{t^8}\cdot(\frac{1}{t^2}+1)}\cdot (-\dfrac{1}{t^2})dt=∫t81⋅(t21+1)1⋅(−t21)dt
=−∫t81+t2dt=−∫(t6−t4+t2−1+11+t2)dt\qquad\qquad =-\int\dfrac{t^8}{1+t^2}dt=-\int(t^6-t^4+t^2-1+\dfrac{1}{1+t^2})dt=−∫1+t2t8dt=−∫(t6−t4+t2−1+1+t21)dt
=−17t7+15t5−13t3+t+arctant+C\qquad\qquad =-\dfrac 17t^7+\dfrac 15t^5-\dfrac 13t^3+t+\arctan t+C=−71t7+51t5−31t3+t+arctant+C
=−17x7+15x5−13x3+1x+arctan1x+C\qquad\qquad =-\dfrac{1}{7x^7}+\dfrac{1}{5x^5}-\dfrac{1}{3x^3}+\dfrac 1x+\arctan \dfrac 1x+C=−7x71+5x51−3x31+x1+arctanx1+C
