IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

srujana

1)

$For\ x>0,\ f(x)=\int_1^x {\frac{lnt}{1+t}}dt.\\\ \\ Find\ the\ value\ of\ f(e)+f\left(\frac{1}{e}\right)$

2)

$If\ f(x)=\int_0^1{\frac{dt}{1+|x-t|},\ then\ f$

Ans fr both the qs = $\frac{1}{2}$

allamraju

1)f(1/x)=

Put k=1/t then dt=(-1/k²)dk.So,the integral becomes

So,f(x)+f(1/x)=

Putting x=e,we get f(e)+f(1/e)=1/2.

RyuAmakusa

i have checked a lot of times the ans for the second q is 0 and not 1/2.i will post the sol. but the editor is not working so it might not be clear. any way put x-t as k so the integral becomes -dk/1+|k| the limits are from x to x-1 u can consume that - to change the limit from x-1 to x now use leibnitz rule u get f'(x) = (1/1+|x-1| ) - (1/1+|x| ) now sub x=1/2 u get the ans to be 0.........

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1)f(1/x)=

Put k=1/t then dt=(-1/k2)dk.So,the integral becomes

So,f(x)+f(1/x)=

Putting x=e,we get f(e)+f(1/e)=1/2.

Put k=1/t then dt=(-1/k²)dk.So,the integral becomes