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For x > 0 let f (x) =integral (1,x) (ln(t)/1+t)dt,find the functionf(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.

janakiram , 11 Years ago
Grade 12
anser 1 Answers
Jitender Singh

Last Activity: 10 Years ago

Ans:
Hello Student,
Please find answer to your question below

f(x) = \int_{1}^{x}\frac{lnt}{1+t}dt
f(\frac{1}{x}) = \int_{1}^{\frac{1}{x}}\frac{lnt}{1+t}dt
t = \frac{1}{z}
dt = \frac{-1}{z^{2}}dz
f(\frac{1}{x}) = \int_{1}^{x}\frac{ln\frac{1}{z}}{1+\frac{1}{z}}.\frac{-dz}{z^{2}}
f(\frac{1}{x}) = \int_{1}^{x}\frac{lnz}{1+z}.\frac{dz}{z}
t, z are both dumy variable which be add together
f(x) + f(\frac{1}{x}) = \int_{1}^{x}(\frac{lnz}{1+z}+\frac{lnz}{z(1+z)})dz
f(x) + f(\frac{1}{x}) = \int_{1}^{x}\frac{lnz}{z}dz
f(x) + f(\frac{1}{x}) =(\frac{ (lnz)^{2}}{2})_{1}^{x}
f(x) + f(\frac{1}{x}) =\frac{ (lnx)^{2}}{2}
f(e) + f(\frac{1}{e}) =\frac{ (lne)^{2}}{2} = \frac{1}{2}


Use Newton – Leibniz theorm, we have
f'(x) = \frac{lnx}{1+x}

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