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For x > 0 let f (x) = integration 1 to x ln(t)/1+t dt, find the function f(x) +f(1/x) and show that f(e)+f(1/e) =1/2 Hint : write f(x) and f(1/x) then use substitution to make their limits same

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6 years ago bharat bajaj
IIT Delhi
122 Points
```							f(x) = (ln t)^2 / 2f(x) = (ln x)^2/2f(x) + f(1/x) = ln x^2 / 2 + (-ln x)^2 / 2 = lnx^2 = 2f(x)Thanksbharat bajajaskiitians facultyiit delhi
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6 years ago
```							F(x)=int 1to x logt/1+t dt F(1/x)=ing 1 to 1/x logt/1+t dt          Now put t=1/z so that the limit can          change to x           =int 1 to x logz/z+1 .1/z          =put z=t          F(x)=ing 1 to x (logt/1+t +logt/1+t.1/t)          =(logt)  ^2/2-0=1/2
```
one year ago
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