Question

∫√x²-9/x dx

Answer 1

令x=3sect
∫[√(x²-9)/x]dx
=∫[√(9sec²t-9)/(3sect)]d(3sect)
=∫[3tan²t·3sect·tant/(3sect)]dt
=∫3(tan³t)dt
=∫[3(1-cos²t)sint/cos³t]dt
=∫[(3/cost)-(3/cos³t)]d(cost)
=3ln|cost|+ (3/2)/cos²t +C
=3ln|3/x| +(3/2)/(3/x)² +C
=-3ln|x/3| +(1/6)x² +C

Answer 2

引用xuzhouliuying的回答：
令x=3sect
∫[√(x²-9)/x]dx
=∫[√(9sec²t-9)/(3sect)]d(3sect)
=∫[3tan²t·3sect·tant/(3sect)]dt
=∫3(tan³t)dt
=∫[3(1-cos²t)sint/cos³t]dt
=∫[(3/cost)-(3/cos³t)]d(cost)
=3ln|cost|+ (3/2)/cos²t +C
=3ln|3/x| +(3/2)/(3/x)² +C
=-3ln|x/3| +(1/6)x² +C

第四行根号开错了