integral (1+x-x^-1)e^x+x^-1 dx=? solution please ….......................
integral (1+x-x^-1)e^x+x^-1 dx=? solution please ….......................
Pavani , 6 Years ago
Grade 12
Integral Calculus> integral (1+x-x^-1)e^x+x^-1 dx=? solution...
1 AnswersArun
Last Activity: 6 Years ago
e same way as for : ∫ Lnx dx = x*Lnx - x
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
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
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