Question

For$x>0$, if$f\left(x\right)={\int }_1^x{\log }_e\frac{t}{(1+t)}dt$, then $f\left(e\right)+f\left(\frac{1}{e}\right)=?$

• A. $1/2$
• B. $-1$
• C. $1$
• D. $0$

Answer 1

The correct option is A

$1/2$

Explanation for the correct option:

Find the value of $\mathit{f}\mathbf{\left(}\mathit{e}\mathbf{\right)}\mathbf{+}\mathit{f}\left(\frac{1}{e}\right)\mathbf{=}\mathbf{?}$ :

Given That,

$f\left(x\right)={\int }_1^x{\log }_e\frac{t}{(1+t)}dt$

So,

$f\left(e\right)+f\left(\frac{1}{e}\right)={\int }_1^e\frac{\ln t}{(1+t)}dt+{\int }_1^{1/e}\frac{\ln t}{(1+t)}dt=I_1+I_2$

$where{I}_1={\int }_1^e\frac{\ln t}{(t+1)}and{I}_2={\int }_1^{1/e}\frac{\ln t}{(t+1)}$

$$\begin{gathered} I_2={\int }_1^{1/e}\frac{\ln t}{(1+t)}dt\mathbf{\left[}\mathbf{P}\mathbf{u}\mathbf{t}\mathbf{}\mathbf{t}\mathbf{=}\frac{\mathbf{1}}{\mathbf{z}}\mathbf{}\mathbf{}\mathbf{\Rightarrow }\mathbf{}\mathbf{d}\mathbf{t}\mathbf{=}\mathbf{-}\frac{\mathbf{d}\mathbf{z}}{{\mathbf{z}}^{\mathbf{2}}}\mathbf{\right]} \\ ={\int }_1^e-\frac{\ln z}{(1+{\displaystyle \frac{1}{z}})}\times \left(-\frac{\mathrm{dz}}{{\mathrm{z}}^2}\right) \\ ={\int }_1^e\frac{\ln z}{z(z+1)}dz \\ ={\int }_1^e\frac{\ln z(z+1-z)}{z(z+1)}dz \\ ={\int }_1^e\left[\frac{\ln z}{z}-\frac{\ln z}{(z+1)}dz\right] \end{gathered}$$

$$\begin{gathered} f\left(e\right)+f\left(\frac{1}{e}\right)={\int }_1^e\frac{\ln t}{(1+t)}dt+{\int }_1^e\frac{\ln t}{t}-{\int }_1^e\frac{\ln t}{(1+t)}dt \\ ={\int }_1^e\frac{\ln t}{t}\mathbf{[}\mathbf{Put}\mathbf{}\mathbf{}\mathbf{l}\mathbf{n}\mathbf{}\mathbf{t}\mathbf{}\mathbf{=}\mathbf{u}\mathbf{}\mathbf{\Rightarrow }\mathbf{}\frac{\mathbf{1}}{\mathbf{t}}\mathbf{d}\mathbf{t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{d}\mathbf{u}\mathbf{]} \\ ={\int }_0^{1}udu={\left[\frac{u^2}{2}\right]}_0^1=\frac{1}{2} \end{gathered}$$

Hence the correct option is (A).