Integral finding

Question

I want to find $$\int \left(1+\frac 1 x \right)\cdot e^{-\frac 1 x} \, dx,$$ with $x>0$. I am thinking to use that if a function $f(x)=e^{g(x)}\cdot g'(x)$ then has the integral $F(x)=e^{g(x)}+c$, but I don't know how exactly. Any ideas?

Answer 1

compute the derivative of $$xe^{-1/x}+C$$

Answer 2

Integrating by part:

$$\int e^{\frac {-1}{x}}dx=x e^{\frac {-1}{x}}-\int \frac {e^{\frac {-1}{x}}}{x}dx$$

Then :

$$\int e^{\frac {-1}{x}}dx+\int \frac {e^{\frac {-1}{x}}}{x}dx=x e^{\frac {-1}{x}}$$

$$\int e^{\frac {-1}{x}}(\frac {1}{x}+1)dx=x e^{\frac {-1}{x}}+K$$

Other way: $$I_1=\int \left(1+\frac 1 x \right)\cdot e^{-\frac 1 x} \, dx= \underbrace{\int e^{\frac {-1}{x}}dx}_\text{Integrate by part}+\int \frac {e^{\frac {-1}{x}}}{x}dx=x e^{\frac {-1}{x}}-\int \frac {e^{\frac {-1}{x}}}{x}dx+\int \frac {e^{\frac {-1}{x}}}{x}dx=xe^{\frac {-1}{x}}+K$$

Answer 3

If $f$ and $g$ are two differentiable functions, recognising that from the product rule $$\frac{d}{dx} \left (e^{f(x)} g(x) \right ) = e^{f(x)} [f'(x) g(x) + g'(x)],$$ one immediately has $$\boxed{\int e^{f(x)} \big{(}f'(x) g(x) + g'(x) \big{)} \, dx = e^{f(x)} g(x) + C}$$

For the integral $\displaystyle{\int \left (1 + \frac{1}{x} \right ) e^{-\frac{1}{x}} \, dx}$ we see $f(x) = -1/x$. Thus $f'(x) = 1/x^2$. The function $g(x)$ can hopefully be found by inspection. We see that $g(x) = x$ giving $g'(x) = 1$.

So for the functions $f(x) = -1/x$ and $g(x) = x$, on applying the result given in the box, one has $$\int \left (1 + \frac{1}{x} \right ) e^{-\frac{1}{x}} \, dx = x e^{-\frac{1}{x}} + C.$$