Question

急！求解下面积分题6.6

Answer 1

6.6
令√[(4-x)/(x-12)]=t，则x=(12t²+4)/(t²+1)
∫√[(4-x)/(x-12)]dx
=∫td[(12t²+4)/(t²+1)]
=(12t²+4)t/(t²+1) -∫[(12t²+4)/(t²+1)]dt
=(12t²+12-8)t/(t²+1) -∫[12 -8/(t²+1)]dt
=12t -8t/(t²+1) -12t +8arctant +C
=8arctan[√[(4-x)/(x-12)] -8√[(4-x)/(x-12)]/[(4-x)/(x-12) +1] +C
=8arctan[√[(4-x)/(x-12)] +√[(4-x)(x-12)] +C

Answer 2

设t=√[(4-x)/(x-12)]

则4-x=t²(x-12)

∴x=(12t²+4)/(t²+1)=12-8/(t²+1)

∴原式=∫t·d[12-8/(t²+1)]

=-∫t·d[8/(t²+1)]

=-8t/(t²+1)+∫8/(t²+1)·dt

=-8t/(t²+1)+8arctant+C

=√[(4-x)(x-12)]+8arctan√[(4-x)/(x-12)]+C

追答

最后一步应该有一个负号，
打掉了。
答案是
-√[(4-x)(x-12)]+8arctan√[(4-x)/(x-12)]+C