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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
one year 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)
one year 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.
one year ago
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