For [, , , N], if [(((x)/(e))^2 x+((e)/(x))^2 x) _e x d x=(1)/()((x)/(e))^ x-(1)/()((e)/(x…

Question

For $$\alpha, \beta, \gamma, \delta \in \mathbb{N}$$, if $$\int\left(\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right) \log _{e} x d x=\frac{1}{\alpha}\left(\frac{x}{e}\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{e}{x}\right)^{\delta x}+C$$

, where $$e=\sum_\limits{n=0}^{\infty} \frac{1}{n !}$$ and $$\mathrm{C}$$ is constant of integration, then $$\alpha+2 \beta+3 \gamma-4 \delta$$ is equal to :

Accepted Answer

We have,

$$\int\left(\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right) \log _{e} x d x=\frac{1}{\alpha}\left(\frac{x}{e}\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{e}{x}\right)^{\delta x}+C$$

$$
\begin{aligned}
\left(\frac{x}{e}\right)^{2 x} & =\left(\frac{e^{\log _e x}}{e}\right)^{2 x} \left[\because x=e^{\log _e x}\right]
\\
& =e^{2\left(x \log _e x-x\right)} .........(i)
\end{aligned}
$$

$$
\text { Also, }\left(\frac{e}{x}\right)^{2 x}=\left(\frac{x}{e}\right)^{-2 x}=e^{-2\left(x \log _e x-x\right)}
$$ .........(ii)

Let,

$$
\begin{aligned}
I & =\int\left[\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right] \log _e x d x \\
& =\int\left[e^{2\left(x \log _e x-x\right)}+e^{-2\left(x \log _e x-x\right)}\right] \log _e x d x ~~~~\text{ [From Eqs. (i) and (ii)] }
\end{aligned}
$$

Let $x \log _e x-x=t$

$$
\log _e x d x=d t
$$

$$
\begin{aligned}
I & =\int\left(e^{2 t}+e^{-2 t}\right) d t=\frac{e^{2 t}}{2}-\frac{e^{-2 t}}{2}+C \\
& =\frac{1}{2}\left(\frac{x}{e}\right)^{2 x}-\frac{1}{2}\left(\frac{e}{x}\right)^{2 x}+C
\end{aligned}
$$

On comparing I with $\frac{1}{\alpha}\left(\frac{x}{e}\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{e}{x}\right)^{\delta x}+C$

$$
\begin{aligned}
& \alpha=2, \beta=2, \gamma=2, \delta=2 \\
& \therefore \alpha+2 \beta+3 \gamma-4 \delta=2+2 \times 2+3 \times 2-4 \times 2=4
\end{aligned}
$$

Solutions · 1