If f α = ∫ 1 α log 10 t 1 + t d t , α > 0 , then f e 3 + f e - 3 is equal to

Given,

 $f\left(\alpha \right)={\int }_1^{\alpha }\frac{{\log }_{10}t}{1+t}dt$

So, $f\left(e^3\right)={\int }_1^{e^3}\frac{\ln t}{\ln 10\left(1+t\right)}\mathrm{dt} \cdots \left(1\right)$

Again $f\left(\alpha \right)={\int }_1^{\alpha }\frac{\ln t}{\left(\ln 10\right)\left(1+t\right)}dt$

Now let $t=\frac{1}{x}\Rightarrow x=\frac{1}{t}$$\Rightarrow dt=\frac{-1}{x^2}dx$

So, $f\left(\alpha \right)={\int }_1^{\frac{1}{a}}\frac{-\ln nx}{\left(\ln 10\right)\left(1+\frac{1}{x}\right)}\left(-\frac{1}{x^2}\right)dx$

$\Rightarrow f\left(\alpha \right)=\frac{1}{l\mathrm{n}10}{\int }_1^{\frac{1}{\alpha }}\frac{lnx}{x\left(x+1\right)}dx$

So, $f\left(e^{-3}\right)=\frac{1}{ln10}{\int }_1^{e^3}\frac{lnt}{t\left(t+1\right)}dt \dots \left(2\right)$

Now adding equation $\left(1\right)$ and $\left(2\right)$ we get,

$f\left(e^3\right)+f\left(e^{-3}\right)$$=\left(\frac{1}{\ln n10}\right){\int }_1^{{\mathrm{c}}^3}\frac{lnt}{\left(1+t\right)}\left[1+\frac{1}{t}\right]dt$

$\Rightarrow f\left(e^3\right)+f\left(e^{-3}\right)=\left(\frac{1}{\ln 10}\right){\int }_1^3\frac{\ln t}{t}dt$

Now let $\ln t=r$$\Rightarrow \frac{dt}{t}=dr$

So, $f\left(e^3\right)+f\left(e^{-3}\right)=\frac{1}{\ln 10}{\int }_0^{3}rdr$

$\Rightarrow f\left(e^3\right)+f\left(e^{-3}\right){\overline{)=\left(\frac{1}{\ln }10\right)\left(\frac{r^2}{2}\right)}}_0^3$

$\Rightarrow f\left(e^3\right)+f\left(e^{-3}\right)=\left(\frac{1}{\log 10}\right)\left(\frac{9}{2}\right)$

$\Rightarrow f\left(e^3\right)+f\left(e^{-3}\right)=\frac{9}{2{\log }_{e}10}$