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integral (1+x-x^-1)e^x+x^-1 dx=? solution please ….......................

integral    (1+x-x^-1)e^x+x^-1    dx=? solution please                 ….......................

Grade:12

1 Answers

Arun
25758 Points
3 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 

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