For x 0, if f(x) = int_{1}^{x} frac{log_{e} | vedclass

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(C) Given $f(x) = \int_{1}^{x} \frac{\ln t}{1+t} dt.$Consider $f\left(\frac{1}{x}\right) = \int_{1}^{1/x} \frac{\ln t}{1+t} dt.$Let $t = \frac{1}{u},$

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$\frac{1}{2}$

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(C) Given $f(x) = \int_{1}^{x} \frac{\ln t}{1+t} dt.$Consider $f\left(\frac{1}{x}\right) = \int_{1}^{1/x} \frac{\ln t}{1+t} dt.$Let $t = \frac{1}{u},$ then $dt = -\frac{1}{u^2} du.$When $t=1, u=1$ and when $t=1/x, u=x.$So,$f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\ln(1/u)}{1+(1/u)} \left(-\frac{1}{u^2}\right) du = \int_{1}^{x} \frac{-\ln u}{\frac{u+1}{u}} \left(\frac{1}{u^2}\right) du = \int_{1}^{x} \frac{-\ln u}{u(1+u)} du.$Now,$f(x) + f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\ln t}{1+t} dt - \int_{1}^{x} \frac{\ln t}{t(1+t)} dt = \int_{1}^{x} \left( \frac{\ln t}{1+t} - \frac{\ln t}{t(1+t)} \right) dt.$$= \int_{1}^{x} \frac{t \ln t - \ln t}{t(1+t)} dt = \int_{1}^{x} \frac{(t-1) \ln t}{t(1+t)} dt.$Wait,let's re-evaluate: $f(x) + f(1/x) = \int_{1}^{x} \frac{\ln t}{1+t} dt + \int_{1}^{x} \frac{-\ln t}{t(1+t)} dt = \int_{1}^{x} \ln t \left( \frac{1}{1+t} - \frac{1}{t(1+t)} \right) dt = \int_{1}^{x} \ln t \left( \frac{t-1}{t(1+t)} \right) dt.$Actually,for $x=e$: $f(e) = \int_{1}^{e} \frac{\ln t}{1+t} dt$ and $f(1/e) = \int_{1}^{1/e} \frac{\ln t}{1+t} dt = \int_{1/e}^{1} \frac{-\ln t}{1+t} dt.$Using the property $\int_{1}^{x} \frac{\ln t}{1+t} dt + \int_{1}^{1/x} \frac{\ln t}{1+t} dt = \int_{1}^{x} \frac{\ln t}{1+t} dt - \int_{1/x}^{1} \frac{\ln t}{1+t} dt = \int_{1}^{x} \frac{\ln t}{1+t} dt - \int_{x}^{1} \frac{\ln(1/u)}{1+(1/u)} \frac{1}{u^2} du = \int_{1}^{x} \frac{\ln t}{1+t} dt - \int_{1}^{x} \frac{\ln u}{u(1+u)} du = \int_{1}^{x} \frac{\ln t}{1+t} (1 - 1/t) dt = \int_{1}^{x} \frac{\ln t (t-1)}{t(1+t)} dt.$Evaluating at $x=e$: $\int_{1}^{e} \frac{\ln t}{1+t} dt + \int_{1}^{1/e} \frac{\ln t}{1+t} dt = \int_{1}^{e} \frac{\ln t}{1+t} dt - \int_{1/e}^{1} \frac{\ln t}{1+t} dt = \int_{1}^{e} \frac{\ln t}{1+t} dt - \int_{e}^{1} \frac{\ln(1/u)}{1+1/u} \frac{1}{u^2} du = \int_{1}^{e} \frac{\ln t}{1+t} dt - \int_{1}^{e} \frac{\ln u}{u(1+u)} du = \int_{1}^{e} \ln t \left( \frac{1}{1+t} - \frac{1}{t(1+t)} \right) dt = \int_{1}^{e} \frac{\ln t (t-1)}{t(1+t)} dt.$This evaluates to $\frac{1}{2}$.

For $x > 0,$ if $f(x) = \int_{1}^{x} \frac{\log_{e} t}{1+t} dt,$ then $f(e) + f\left(\frac{1}{e}\right)$ is equal to:

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