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For `x >0,l e tf(x)=int_1^x(logt)/(1+t)dtdot`Find the function `f(x)+f(1/x)`and find the value of `f(e)+f(1/e)dot` |
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Answer» Let `F(x)=(x)+f(1/x)` `=int_(1)^(x)(logt)/(1+t)dt+int_(1)^(1//x)(logt)/(1+t)dt` In second integral let `t=1//y`. So `dt=-1/(y^(2))dy` `:.F(x)=int_(1)^(x)(logt)/(1+t)dt+int_(1)^(x)(-logy)/(1+1/y)(-(dy)/(y^(2)))` `=int_(1)^(x)(logt)/(1+t)dt+int_(1)^(x)(logy)/(y(1+y))dy` `=int_(1)^(x)(logt)/(1+t)dt+int_(1)^(x)(logt)/(t(1+t))dt` `=int_(1)^(x)(logt)/t dt=1/2(logx)^(2)` `:.F(e)=1/2` |
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