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find the integration of (1+x-1/x)^e^(x+1/x) options are: 1.(x-1)e^(x+1/x)+c 2.xe^(x+1/x)+c

find the integration of (1+x-1/x)^e^(x+1/x)
options are:
1.(x-1)e^(x+1/x)+c
2.xe^(x+1/x)+c

Question Image
Grade:12th pass

2 Answers

Arun
25763 Points
3 years ago
Dear student
 
ok, fair enough. 

the 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 
 
hence option 2 is correct.
 
Regards
Arun (askIITians forum expert)
abhijeet
11 Points
3 years ago
 Let u = x+1/x.
Then du = (1-1/x^2) dx.
So
integral(1+x-1/x)e^(x+1/x) dx = integral [1+x(1-1/x^2)]e^u dx
= integral [e^u dx + xe^u du]
= integral [d(xe^u)]
= xe^u+C.

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