第三题，求不定积分
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let
y= (1+x^2)
dy = 2xdy
∫ 6x/[1+ (1+x^2)^(1/3)] dx
=∫ 3/[1+ y^(1/3)] dy
let
u= y^(1/3)
3u^2 .du = dy
∫ 6x/[1+ (1+x^2)^(1/3)] dx
=∫ 3/[1+ y^(1/3)] dy
=9∫ u^2/(1+ u) du
=9∫ [(u-1) + 1/(1+ u)] du
=9 [ (1/2)u^2 - u + ln|1+u| ] + C
=9 [ (1/2)y^(2/3) - y^(1/3) + ln|1+y^(1/3)| ] + C
=9 [ (1/2)(1+x^2)^(2/3) - (1+x^2)^(1/3) + ln|1+(1+x^2)^(1/3)| ] + C
y= (1+x^2)
dy = 2xdy
∫ 6x/[1+ (1+x^2)^(1/3)] dx
=∫ 3/[1+ y^(1/3)] dy
let
u= y^(1/3)
3u^2 .du = dy
∫ 6x/[1+ (1+x^2)^(1/3)] dx
=∫ 3/[1+ y^(1/3)] dy
=9∫ u^2/(1+ u) du
=9∫ [(u-1) + 1/(1+ u)] du
=9 [ (1/2)u^2 - u + ln|1+u| ] + C
=9 [ (1/2)y^(2/3) - y^(1/3) + ln|1+y^(1/3)| ] + C
=9 [ (1/2)(1+x^2)^(2/3) - (1+x^2)^(1/3) + ln|1+(1+x^2)^(1/3)| ] + C
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