Question

2,3,5题怎么做，详细解答哦

Answer 1

(2)
令√(x-1)=t，则x=t²+1
∫1/[1+√(x-1)]dx
=∫[1/(1+t)]d(t²+1)
=∫[2t/(1+t)]dt
=∫[(2t+2-2)/(1+t)]dt
=∫[2 -2/(1+t)]dt
=2t -2ln|1+t| +C
=2√(x-1) -2ln|1+√(x-1)| +C
(3)
∫[(2x+1)/(x²+3x-10)]dx
=(1/7)∫[5/(x-2)+ 9/(x+5)]dx
=(5/7)ln|x-2| +(9/7)ln|x+5| +C
(5)
√arctan(1/x)=t，则x=cot(t²)
∫[√arctan(1/x)/(1+x²)]dx
=∫t/[1+cot²(t²)]d[cot(t²)]
=-∫2t²·csc(t²)/[1+cot²(t²)]dt
=-∫2t²·csc(t²)/[1+cot²(t²)]dt
=-∫2t²dt
=-⅔t³ +C
=-⅔[arctan(1/x)]^(3/2) +C