Question

The value of the integral $\int\limits^2_{1}e^{x} \left(log_{e} \,x + \frac{x+1}{x}\right)dx$ is

• A. $e^2(1 + log_e 2)$
• B. $e^2 - e$
• C. $e^2(1 + log_e 2) - e$
• D. $e^2 -e(1 + log_e 2)$

Answer 1

Answer: (C)

$e^2(1 + log_e 2) - e$

Answer 2

Let $/=\int_{1}^{2} e^{x}\left(\log _{\theta} x+\frac{x+1}{x}\right) d x$
\Rightarrow I=\int_\limits{1}^{2}\left(e^{x} \cdot \log _{e} x+e^{x}+\frac{e^{x}}{x}\right) d x
\Rightarrow I=\int_\limits{1}^{2} e^{x} \log _{\theta} x d x+\int_\limits{1}^{2} e^{x} d x+\int_\limits{1}^{2} \frac{e^{x}}{x} d x
\Rightarrow I=\int_\limits{1}^{2} e^{x} \log _{e} x d x+\left[e^{x}\right]_{1}^{2}+\left[e^{x} \log _{\theta} x_{1}^{2}\right.
-\int_\limits{1}^{2} e^{x} \log _{\theta} x d x
$\Rightarrow I=\left(e^{2}-e^{1}\right)+\left(e^{2} \log _{\theta} 2-0\right)$
$=e^{2}\left(1+\log _{\theta}\right.$ 2) $-e$