Question

Find the value of the integration $\int {(1 + x - \dfrac{1}{x}){e^{\left( {x + \frac{1}{x}} \right)}}dx} $ .
A.$(x - 1){e^{\left( {x + \frac{1}{x}} \right)}} + c$
B.$x{e^{\left( {x + \frac{1}{x}} \right)}} + c$
C.$(x + 1){e^{\left( {x + \frac{1}{x}} \right)}} + c$
D. $ - x{e^{\left( {x + \frac{1}{x}} \right)}} + c$

Answer 1

Hint: First rewrite the given integrant as $\int {1 + x\left( {1 - \dfrac{1}{{{x^2}}}} \right){e^{\left( {x + \frac{1}{x}} \right)}}dx} $ , then split the integration. After that integrate the first part with the help of by parts integration and keep the second part as it is.

Formula Used:
The formula of by parts integration is,
$\int {uvdx = u\int {vdx - \int {\left[ {\dfrac{d}{{dx}}(u)\int {vdx} } \right]} dx} } $ , where u and v are the functions of x only.

Complete step by step solution:
Rewrite the give integration as,
$\int {\left[ {1 + x\left( {1 - \dfrac{1}{{{x^2}}}} \right){e^{\left( {x + \frac{1}{x}} \right)}}} \right]dx} $.
So, $\int {\left[1 + x\left( {1 - \dfrac{1}{{{x^2}}}} \right){e^{\left( {x + \frac{1}{x}} \right)}}\right]dx} = \int {{e^{\left( {x + \frac{1}{x}} \right)}}} dx + \int {x\left( {1 - \dfrac{1}{{{x^2}}}} \right){e^{\left( {x + \frac{1}{x}} \right)}}dx} $---(1)
Now, integrate $\int {{e^{\left( {x + \frac{1}{x}} \right)}}} dx$ with by parts rule.
$\int {{e^{\left( {x + \frac{1}{x}} \right)}}} dx = {e^{\left( {x + \frac{1}{x}} \right)}}\int {dx - \int {\left[ {\dfrac{d}{{dx}}{e^{\left( {x + \frac{1}{x}} \right)}}\int {dx} } \right]dx} } $
$ = x{e^{\left( {x + \frac{1}{x}} \right)}} - \int {{e^{\left( {x + \frac{1}{x}} \right)}}\dfrac{d}{{dx}}\left( {x + \dfrac{1}{x}} \right)xdx} $
$ = x{e^{\left( {x + \frac{1}{x}} \right)}} - \int {x\left( {1 - \dfrac{1}{{{x^2}}}} \right){e^{\left( {x + \frac{1}{x}} \right)}}dx} + c$--(2), where c is the integrating constant.
From (1) and (2) we have,
$\int {\left[ {1 + x\left( {1 - \dfrac{1}{{{x^2}}}} \right){e^{\left( {x + \frac{1}{x}} \right)}}} \right]dx} = x{e^{\left( {x + \frac{1}{x}} \right)}} - \int {x\left( {1 - \dfrac{1}{{{x^2}}}} \right){e^{\left( {x + \frac{1}{x}} \right)}}dx} + \int {x\left( {1 - \dfrac{1}{{{x^2}}}} \right){e^{\left( {x + \frac{1}{x}} \right)}}dx} + c$
$ = x{e^{\left( {x + \frac{1}{x}} \right)}} + c$

Option ‘B’ is correct

Note: Do not use the by part formula in both parts of the integral (1) that is not required, you just need to integrate the first part; the second one will get canceled at the end.