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The value of the integral ∫(1+x-1/x)e x+1/x is equal to=

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2 years ago

```							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  RegardsArun
```
2 years ago
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