The value of the integral ∫(1+x-1/x)ex+1/x is equal to=
The value of the integral ∫(1+x-1/x)ex+1/x is equal to=
Shayak Jana , 7 Years ago
Grade 12th pass
Integral Calculus> The value of the integral ∫(1+x-1/x)e x+1...
1 Answers
Arun
Dear student
we notice that :
(x + 1/x) ' = 1 - 1/x^2
and so :
x (x + 1/x) ' = x (1 - 1/x^2) = x - 1/x
so for ----> u(x) = x+ 1/x
we have to integrate a form that can be presented like this:
(1+x-1/x)*e^(x+1/x) = 1*e^u(x) + x u '(x) *e^u(x) can be written like this :
(x) ' *e^u(x) + x u '(x) *e^u(x)
which is the derivative of ------> [ x e^u(x) ]
and therefore we see that :
∫ (1+x-1/x)*e^(x+1/x) dx = ∫ [ (x) ' *e^u(x) + x u '(x) *e^u(x) ] dx
= ∫ [ (x) ' *e^u(x) + x [e^u(x)] ' ] dx
= ∫ [ x e^u(x) ] ' dx
= x e^u(x) + C
conclusion :
∫ (1+x-1/x)*e^(x+1/x) dx = x e^(x+1/x) + C
(x + 1/x) ' = 1 - 1/x^2
and so :
x (x + 1/x) ' = x (1 - 1/x^2) = x - 1/x
so for ----> u(x) = x+ 1/x
we have to integrate a form that can be presented like this:
(1+x-1/x)*e^(x+1/x) = 1*e^u(x) + x u '(x) *e^u(x) can be written like this :
(x) ' *e^u(x) + x u '(x) *e^u(x)
which is the derivative of ------> [ x e^u(x) ]
and therefore we see that :
∫ (1+x-1/x)*e^(x+1/x) dx = ∫ [ (x) ' *e^u(x) + x u '(x) *e^u(x) ] dx
= ∫ [ (x) ' *e^u(x) + x [e^u(x)] ' ] dx
= ∫ [ x e^u(x) ] ' dx
= x e^u(x) + C
conclusion :
∫ (1+x-1/x)*e^(x+1/x) dx = x e^(x+1/x) + C
Regards
Arun
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