If f(alpha) = int_{1}^{alpha} frac{log_{10} t | vedclass

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(D) Given $f(\alpha) = \int_{1}^{\alpha} \frac{\ln t}{(\ln 10)(1+t)} dt$.$f(e^{3}) = \int_{1}^{e^{3}} \frac{\ln t}{(\ln 10)(1+t)} dt \quad \dots(1)$Fo

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$\frac{9}{2 \log_{e}(10)}$

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(D) Given $f(\alpha) = \int_{1}^{\alpha} \frac{\ln t}{(\ln 10)(1+t)} dt$.$f(e^{3}) = \int_{1}^{e^{3}} \frac{\ln t}{(\ln 10)(1+t)} dt \quad \dots(1)$For $f(e^{-3}) = \int_{1}^{e^{-3}} \frac{\ln t}{(\ln 10)(1+t)} dt$,let $t = \frac{1}{x}$,so $dt = -\frac{1}{x^{2}} dx$.When $t=1, x=1$ and when $t=e^{-3}, x=e^{3}$.$f(e^{-3}) = \int_{1}^{e^{3}} \frac{\ln(1/x)}{(\ln 10)(1+1/x)} \left(-\frac{1}{x^{2}}\right) dx = \int_{1}^{e^{3}} \frac{-\ln x}{(\ln 10)(\frac{x+1}{x})} \left(-\frac{1}{x^{2}}\right) dx = \frac{1}{\ln 10} \int_{1}^{e^{3}} \frac{\ln x}{x(x+1)} dx \quad \dots(2)$Adding $(1)$ and $(2)$:$f(e^{3}) + f(e^{-3}) = \frac{1}{\ln 10} \int_{1}^{e^{3}} \left( \frac{\ln t}{1+t} + \frac{\ln t}{t(1+t)} \right) dt$$= \frac{1}{\ln 10} \int_{1}^{e^{3}} \frac{\ln t}{1+t} \left( 1 + \frac{1}{t} \right) dt = \frac{1}{\ln 10} \int_{1}^{e^{3}} \frac{\ln t}{1+t} \left( \frac{t+1}{t} \right) dt$$= \frac{1}{\ln 10} \int_{1}^{e^{3}} \frac{\ln t}{t} dt$Let $\ln t = u$,then $\frac{1}{t} dt = du$. When $t=1, u=0$ and when $t=e^{3}, u=3$.$= \frac{1}{\ln 10} \int_{0}^{3} u du = \frac{1}{\ln 10} \left[ \frac{u^{2}}{2} \right]_{0}^{3} = \frac{1}{\ln 10} \left( \frac{9}{2} \right) = \frac{9}{2 \log_{e} 10}$.

If $f(\alpha) = \int_{1}^{\alpha} \frac{\log_{10} t}{1+t} dt, \alpha > 0$,then $f(e^{3}) + f(e^{-3})$ is equal to.

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