前置知识：第二类换元法
计算 ∫ 1 x 8 ( x 2 + 1 ) d x \int\dfrac{1}{x^8(x^2+1)}dx ∫x8(x2+1)1dx
解：
\qquad
令
x
=
1
t
x=\dfrac 1t
x=t1，
t
=
1
x
t=\dfrac 1x
t=x1，
d
x
=
−
1
t
2
d
t
dx=-\dfrac{1}{t^2}dt
dx=−t21dt
\qquad 原式 = ∫ 1 1 t 8 ⋅ ( 1 t 2 + 1 ) ⋅ ( − 1 t 2 ) d t =\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
= − ∫ t 8 1 + t 2 d t = − ∫ ( t 6 − t 4 + t 2 − 1 + 1 1 + t 2 ) d t \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
= − 1 7 t 7 + 1 5 t 5 − 1 3 t 3 + t − arctan t + C \qquad\qquad =-\dfrac 17t^7+\dfrac 15t^5-\dfrac 13t^3+t-\arctan t+C =−71t7+51t5−31t3+t−arctant+C
= − 1 7 x 7 + 1 5 x 5 − 1 3 x 3 + 1 x − arctan 1 x + 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
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