dx/x(1-ln^2 x))^1/2 from 1 to e
and
(e^1/x)dx/x^2 from 1 to 2
and
(e^1/x)dx/x^2 from 1 to 2
-
∫dx /x√1 - (ln x)^2] from 1 to e
let ln x = u, when x = 1, u = 0 and when x = e, u = 1
dx/x = du
∫ du /√(1 - u^2) from 0 to 1
= sin^-1(u) from 0 to 1
= Π/2 - 0
= Π/2
2)
∫e^(1/x)dx /x^2 from 1 to 2
let 1/x = u, when x = 1, u = 1 and when x = 2, u = 1/2
-dx/x^2 = du
dx/x^2 = -du
- ∫e^(u) du from 1 to 1/2
= -e^u from 1 to 1/2
= - [e^(1/2) - e^1 ]
= e - e^(1/2)
= 1.0696
let ln x = u, when x = 1, u = 0 and when x = e, u = 1
dx/x = du
∫ du /√(1 - u^2) from 0 to 1
= sin^-1(u) from 0 to 1
= Π/2 - 0
= Π/2
2)
∫e^(1/x)dx /x^2 from 1 to 2
let 1/x = u, when x = 1, u = 1 and when x = 2, u = 1/2
-dx/x^2 = du
dx/x^2 = -du
- ∫e^(u) du from 1 to 1/2
= -e^u from 1 to 1/2
= - [e^(1/2) - e^1 ]
= e - e^(1/2)
= 1.0696
-
1) INTEGRAL[dx/[x(1-ln^2(x))^1/2]: 1 to e]
u=ln(x) then du=dx/x
INTEGRAL[1/sqrt(1-u^2) du]
sin-1(u)+C
sin-1(ln(x)):1 to e
=sin-1(1)=pi/2
#2) INTEGRAL[e^(1/x)/x^2 dx:1 to 2]
u=1/x and du=-1/x^2 then -du=1/x^2
INTEGRAL[-e^u du]
-e^u+C
-e^(1/x):1 to 2
-e^(1/2)+e = e-sqrt(e)
u=ln(x) then du=dx/x
INTEGRAL[1/sqrt(1-u^2) du]
sin-1(u)+C
sin-1(ln(x)):1 to e
=sin-1(1)=pi/2
#2) INTEGRAL[e^(1/x)/x^2 dx:1 to 2]
u=1/x and du=-1/x^2 then -du=1/x^2
INTEGRAL[-e^u du]
-e^u+C
-e^(1/x):1 to 2
-e^(1/2)+e = e-sqrt(e)
-
plug, use fnInt in your calculator