Mathematics

# Evaluate: $\displaystyle \int (1+x -x^{-1})e^{x+x^{-1}}dx$

$xe^{x+x^{-1}} + c$

##### SOLUTION
We know that;
$\int { { e }^{ x }\left( f(x)+f'(x) \right) dx } ={ e }^{ x }f\left( x \right) +C$
But there's a more generalized form which deals with functions of $x$ in the exponent;
$\int { { e }^{ g\left( x \right) }\left( f(x)g'(x)+f'(x) \right) dx } ={ e }^{ g\left( x \right) }f\left( x \right) +C$
The above integral has been derived by differentiating ${ e }^{ g\left( x \right) }f\left( x \right)$, similar to the derivation of the formula, $\int { { e }^{ x }\left( f(x)+f'(x) \right) dx } ={ e }^{ x }f\left( x \right) +C$.
Hence,
$I=\int { \left( 1+x-\cfrac { 1 }{ x } \right) { e }^{ x+\tfrac { 1 }{ x } }dx } \\ I=\int { \left[ x\left( 1-\cfrac { 1 }{ { x }^{ 2 } } \right) +1 \right] { e }^{ x+\tfrac { 1 }{ x } }dx }$
Over here;
$f\left( x \right) =x\\ g\left( x \right) =x+\cfrac { 1 }{ x } \\ f'\left( x \right) =1\\ g^{ ' }\left( x \right) =1-\cfrac { 1 }{ { x }^{ 2 } }$
Therefore the integral is in the form of;
$\int { { e }^{ g\left( x \right) }\left( f(x)g'(x)+f'(x) \right) dx } ={ e }^{ g\left( x \right) }f\left( x \right) +C$
$I=x{ e }^{ \left( x+\tfrac { 1 }{ x } \right) }+C$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

#### Realted Questions

Q1 Subjective Hard
Prove that $\displaystyle \int \sqrt{x^2 - a^2} dx = \dfrac{x}{2} \sqrt{x^2 - a^2} - \dfrac{a^2}{2} \log |x + \sqrt{x^2 - a^2}| + c$.

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\displaystyle \int \frac {dx}{e^x+e^{-x}}$ is equal to
• A. $\tan^{-1}(e^{-x})+C$
• B. $\log (e^x-e^{-x})+C$
• C. $\log (e^x+e^{-x})+C$
• D. $\tan^{-1}(e^x)+C$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
If $\displaystyle\int \dfrac{d\theta}{\cos^{2}\theta(\tan 2\theta+\sec 2\theta)}=$
$\lambda \tan\theta+2\log_{e}|f(\theta)|+C$ where $C$ is a constant of integrating, then the ordered pair $(\lambda, f(\theta))$ is equal to:
• A. $(1,1-\tan\theta)$
• B. $(-1,1-\tan\theta)$
• C. $(-1,1+\tan\theta)$
• D. $(1,1+\tan\theta)$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\displaystyle \int x^{3}e^{x}dx=$
• A. $e^{x}(x^{3}-3x^{2}+6x+6)+c$
• B. $e^{x}(x^{3}+3x^{2}+6x+6)+c$
• C. $e^{x}(x^{3}- 3x^{2}-6x+6)+c$
• D. $e^{x}(x^{3}- 3x^{2}+6x-6)+c$

$\displaystyle\int x^n\log_ex\ dx$