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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)+c2.xe^(x+1/x)+c`
2 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. RegardsArun (askIITians forum expert)
```
2 years ago
```							 Let u = x+1/x.Then du = (1-1/x^2) dx.Sointegral(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.
```
2 years ago
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