Question

If $f(\alpha )=\underset{1}{\overset{\alpha }{\int }}\frac{{\log }_{10}t}{1+t}dt,\alpha >0$, then $f\left(e^3\right)+f$ $\left(e^{-3}\right)$ is equal to :

• A. 9
• B. $\frac{9}{2}$
• C. $\frac{9}{{\log }_e(10)}$
• D. $\frac{9}{2{\log }_c(10)}$

Answer 1

Answer: (D)

$f\left(e^3\right)=\underset{1}{\overset{e^3}{\int }}\frac{\ell nt}{\ln 10(1+t)}dt\dots ...(1)$
$f(\alpha )=\underset{1}{\overset{a}{\int }}\frac{\ell nt}{(\ln 10)(1+t)}dt$
$t=\frac{1}{x}\Rightarrow x=\frac{1}{t}$
$dt=\frac{-1}{x^2}dx$
$=\underset{1}{\overset{\frac{1}{\alpha }}{\int }}\frac{-\ell \ln x}{(\ell \ln 10)\left(1+\frac{1}{x}\right)}\left(-\frac{1}{x^2}\right)dx$
$f(\alpha )=\frac{1}{\ell \ln 10}\underset{1}{\overset{\frac{1}{\alpha }}{\int }}\frac{\ell nx}{x(x+1)}dx$
$f\left(e^{-3}\right)=\frac{1}{\ell \ln 10}\underset{1}{\overset{c^3}{\int }}\frac{\ell nt}{t(t+1)}dt\dots \dots \dots$
Add (1) & (2)
$f\left(e^3\right)+f\left(e^{-3}\right)$
$=\left(\frac{1}{\ell n10}\right)\underset{1}{\overset{c^3}{\int }}\frac{\ell nt}{(1+t)}\left[1+\frac{1}{t}\right]dt$
$=\left(\frac{1}{\ell n10}\right)\underset{1}{\overset{3}{\int }}\frac{\ell nt}{t}dt$
$\ell nt=r$
$\frac{dt}{t}=dr$
$=\frac{1}{\ell n10}\underset{0}{\overset{3}{\int }}rdr$
$={\left(\frac{1}{\ell n10}\right)\left(\frac{r^2}{2}\right)|}_0^3$
$=\left(\frac{1}{\log 10}\right)\left(\frac{9}{2}\right)$
$=\frac{9}{2{\log }_{e}10}$