3道高数大一上,

(3)
∫ dx/[x.√(x^2-1)]
let
x= secy
dx= (secy)(tany) dy
∫ dx/[x.√(x^2-1)]
=∫ dy
=y + C
= arcsec(x) + C
(5)
∫(2->3) {(x-3)/[(x-1)(x^2-1)] } dx
let
(x-3)/[(x-1)(x^2-1)] ≡ A1/(x-1) + A2/(x-1)^2 + B/(x+1)
x-3 ≡ A1(x-1)(x+1) + A2(x+1) + B(x-1)^2
put x=1
2A2 = -2
A2 =-1
put x=-1
4B= -4
B=-1
coef.of x^2
A1 +B = 0
A1 = 1
ie
(x-3)/[(x-1)(x^2-1)] ≡ 1/(x-1) - 1/(x-1)^2 - 1/(x+1)
∫(2->3) {(x-3)/[(x-1)(x^2-1)} dx
=∫(2->3) [ 1/(x-1) - 1/(x-1)^2 - 1/(x+1) ] dx
=[ ln( (x-1)/(x+1) ) + (x-1) ](2->3)
= 2 + ln(1/2) - 1 - ln( 1/3)
=1+ ln(3/2)
(five)
f(1)=1,f(2)=2
∫(0->x)(2x-t)f(t) dt = 5x^3+1
To find :∫(1->2) f(x) dx
let
f(t)dt = dF(t)
∫(0->x)(2x-t)f(t) dt
=∫(0->x)(2x-t)dF(t)
=[(2x-t)F(t)](0->x) + ∫(0->x) F(t) dt
= xF(x) - 2xF(0) + ∫(0->x) F(t) dt
∫(0->x)(2x-t)f(t) dt = 5x^3+1
xF(x) - 2xF(0) + ∫(0->x) F(t) dt = 5x^3+1
[xF(x) - 2xF(0) + ∫(0->x) F(t) dt ]'= [5x^3+1]'
xf(x) + F(x) - 2F(0) +F(x) = 15x^2
xf(x) + 2F(x) - 2F(0) = 15x^2
put x=1
1f(1)+ 2F(1) -2F(0) = 15
F(1) -F(0) =7 (1)
put x=2
2f(2) + 2F(2) - 2F(0) = 60
F(2)-F(0) = 28 (2)
(2)-(1)
F(2) -F(1) = 21
∫(1->2)f(t) dt
=F(2) -F(1)
=21