Question

高数不定积分计算～～

Answer 1

1. 设 x + 2 = t³，则 dx = 3t² * dt。那么，这个积分就可以变换为：

   =∫3t² *dt/(1+t)

   =3∫(t²-1 +1)*dt/(1+t)

   =3∫[(t-1) + 1/(1+t)] *dt

   =3[∫t*dt - ∫dt + ∫dt/(1+t)]

   =3[t²/2 - t + ln(1+t)] + C

   =3[³√(x+2)² - ³√(x+2) + ln[1+³√(x+2)} + C

2. 设 x = t^6，则 dx = 6t^5 * dt。那么，上面的积分就可以变换为：

   =∫6t^5 * dt/[(1+t²)*t³]

   =6∫[t²/(1+t²)]*dt

   =6∫[1 - 1/(1+t²)]*dt

   =6[∫dt - ∫dt/(1+t²)]

   =6[t - arctan(t)] + C

   =6[x^(-6) - arctan(x^-6)] + C

追问

数算错了，思路是对的

Answer 2

1. 令（x+2)开三次方=t

2. 令x开六次方=t，