Evaluate: `int_0^(e-1)((x^2+2x-1)/2)/(x+1)dx+int_1^e xlogx e^(x^(2-2)/2)dx`

Answer

`e^((e^(2)-2)/2)`

Explanation

Put `x+1=t` in first integral. Then
`int_(1)^(e)(e^((t^(2)-2)/2))/tdt+int_(1)^(e)x log xe^((x^(2)-2)/2)dx`
`int_(1)^(e)(e^((x^(2)-2)/2))/x dt+int_(1)^(e)x log x e^((x^(2)-2)/2)dx`
`=[e^((x^(2)-2)/2)logx]_(1)^(e)-int_(1)^(e)xe^((x^(2)-2)/2) log x dx +int_(1)^(e)x log x e^((x^(2)-2)/2)dx`
`=e^((e^(2)-2)/2)`