求不定积分,用换元法!1)∫1/根号(x^2+1)^3 dx2)∫1/根号x+立方根号x dx

来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/23 22:13:02
求不定积分,用换元法!1)∫1/根号(x^2+1)^3dx2)∫1/根号x+立方根号xdx求不定积分,用换元法!1)∫1/根号(x^2+1)^3dx2)∫1/根号x+立方根号xdx求不定积分,用换元法

求不定积分,用换元法!1)∫1/根号(x^2+1)^3 dx2)∫1/根号x+立方根号x dx
求不定积分,用换元法!
1)∫1/根号(x^2+1)^3 dx
2)∫1/根号x+立方根号x dx

求不定积分,用换元法!1)∫1/根号(x^2+1)^3 dx2)∫1/根号x+立方根号x dx
1) 令:x=tant ,√(x^2+1)^3 = sec³t ,cost = 1/√(x^2+1) ,dx = sec²t dt
∫1/√(x^2+1)^3 dx
=∫1/sec³t * (sec²t dt)
=∫cost dt
= sint + C
= tant*cost + C
= x/√(x^2+1) + C
2)令:x=t^6 ,
∫1/[√x + ³√x ] dx
=∫1/[t² + t³] (6t^5 dt)
= 6*∫t^3/[1 + t] dt
= 6*∫[(t^3+1)-1]/[1 + t] dt
= 6*∫[(t^2 - t +1) -1/(1+t) ]dt
= 2*t^3 - 3*t^2 + 6*t -6*ln(1+t) + C
= 2*√x - 3*³√x + 6*x^(1/6) - 6*ln(1+x^(1/6)) + C