第二类换元法之幂代换习题
前置知识：第二类换元法
计算 ∫ 1 x + x 3 d x \int \dfrac{1}{\sqrt x+\sqrt[3]{x}}dx ∫x+3x1dx
解：
\qquad
令
x
=
t
6
x=t^6
x=t6，
d
x
=
6
t
5
d
t
dx=6t^5dt
dx=6t5dt
\qquad 原式 = ∫ 6 t 5 t 3 + t 2 d t = ∫ 6 t 3 t + 1 d t =\int\dfrac{6t^5}{t^3+t^2}dt=\int\dfrac{6t^3}{t+1}dt =∫t3+t26t5dt=∫t+16t3dt
= 6 ∫ t 2 ( t + 1 ) − t ( t + 1 ) + ( t + 1 ) − 1 t + 1 d t \qquad\qquad =6\int \dfrac{t^2(t+1)-t(t+1)+(t+1)-1}{t+1}dt =6∫t+1t2(t+1)−t(t+1)+(t+1)−1dt
= 6 ∫ ( t 2 − t + 1 − 1 t + 1 ) d t = 6 ( 1 3 t 3 − 1 2 t 2 + t − ln ∣ t + 1 ∣ ) + C \qquad\qquad =6\int(t^2-t+1-\dfrac{1}{t+1})dt=6(\dfrac 13t^3-\dfrac 12t^2+t-\ln|t+1|)+C =6∫(t2−t+1−t+11)dt=6(31t3−21t2+t−ln∣t+1∣)+C
= 2 t 3 − 3 t 2 + 6 t − 6 ln ∣ t + 1 ∣ + C = 2 x − 3 x 3 + 6 x 6 − ln ∣ x 6 + 1 ∣ + C \qquad\qquad =2t^3-3t^2+6t-6\ln|t+1|+C=2\sqrt{x}-3\sqrt[3]{x}+6\sqrt[6]{x}-\ln|\sqrt[6]{x}+1|+C =2t3−3t2+6t−6ln∣t+1∣+C=2x−33x+66x−ln∣6x+1∣+C