Mathematics

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


ANSWER

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


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;
$$\int { { e }^{ g\left( x \right)  }\left( f(x)g'(x)+f'(x) \right) dx } ={ e }^{ g\left( x \right)  }f\left( x \right) +C$$
hence the answer is;
$$I=x{ e }^{ \left( x+\tfrac { 1 }{ x }  \right)  }+C$$
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