Question

求不定积分，∫1/(x^(1/2)+x^(1/3)) dx

答案是2(x^1/2)-3(x^1/3)+6(x^1/6)-6ln(1+x^1/6)+C,注意，最后一个括号是绝
对值号。

Answer 1

这就是一个很简单的三角换元，令x=sint，则dx=costdt,∫(1-x^2)^(3/2)
dx=∫cost(1-(sint)^2)^3/2dt=∫(cost)^4dt=∫((cos4t)/8+(cos2t)/2+3/8)dt(二倍角公式得到的)=-(sin4t)/32-(sin2t)/4+3t/8
=-sintcost(1-2(sint)^2)/8-sintcost/2+3t/3(还是二倍角)=-x(1-x^2)^1/2(5-2x^2)/8+3arcsinx/8

Answer 2

x=t^6,t=x^(1/6),dx=6t^5 dt
∫1/(x^(1/2)+x^(1/3)) dx=∫1/(t^3+t^2) *6t^5 dt
=S6t^3/(t+1)dt
=6S(t^3+t^2-t^2-t+t+1-1)(t+1) dt
=6S(t^2-t+1)dt-6S1/(t+1)d(t+1)
=2t^3-3t^2+6t-6ln|t+1|+c
=2(x^1/2)-3(x^1/3)+6(x^1/6)-6ln(1+x^1/6)+C

Answer 3

∫dx/(x^(1/2)+x^(1/3))
=∫dx/[x^(1/3)(x^(1/6)+1)]
=(3/2)∫dx^(2/3)/[x^(1/6)+1]
=(3/2)∫4x^(3/6)dx^(1/6)/[x^(1/6)+1]
=6∫[x^(3/6)+x^(2/6)]dx^(1/6)/[x^(1/6)+1] -6∫[x^(2/6)+x(1/6)]dx^(1/6)/[x^(1/6)+1]
+6∫[x^(1/6)+1]dx/[x^(1/6)+1] -6∫d[x^(1/6)+1]/[x^(1/6)+1]
=6∫x^(2/6)dx^(1/6) - 6∫x^(1/6)dx^(1/6) +6∫dx^(1/6) -6ln|x^(1/6)+1|
=2x^(3/6)-3x^(2/6)+6x^(1/6)-6ln(x^(1/6)+1)+C