# The integral

Question:

The integral $\int_{1}^{e}\left\{\left(\frac{x}{e}\right)^{2 x}-\left(\frac{e}{x}\right)^{x}\right\} \log _{e} x d x$ is equal to :

1. (1) $\frac{1}{2}-e-\frac{1}{e^{2}}$

2. (2) $-\frac{1}{2}+\frac{1}{e}-\frac{1}{2 e^{2}}$

3. (3) $\frac{3}{2}-\frac{1}{e}-\frac{1}{2 e^{2}}$

4. (4) $\frac{3}{2}-e-\frac{1}{2 e^{2}}$

Correct Option: , 4

Solution:

$I=\int_{1}^{e}\left\{\left(\frac{x}{e}\right)^{2 x}-\left(\frac{e}{x}\right)^{x}\right\} \log _{e} x d x$

Let $\left(\frac{x}{e}\right)^{x}=t$

$\Rightarrow \quad x \ln \left(\frac{x}{e}\right)=\ln t$

$\Rightarrow \quad x(\ln x-1)=\ln t$

On differentiating both sides w.r. $t x$ we get

$\ln x \cdot d x=\frac{d t}{t}$

When $x=e$ then $t=1$ and when $x=1$ then $t=\frac{1}{e}$.

$I=\int_{\frac{1}{e}}^{1}\left(t^{2}-\frac{1}{t}\right) \cdot \frac{d t}{t}=\int_{\frac{1}{e}}^{1}\left(t-\frac{1}{t^{2}}\right) d t$

$=\left(\frac{t^{2}}{2}+\frac{1}{t}\right)_{\frac{1}{e}}^{1}=\left(\frac{1}{2}+1\right)-\left(\frac{1}{2 e^{2}}+e\right)=\frac{3}{2}-e-\frac{1}{2 e^{2}}$