Integrate \frac{1}{x}-\frac{1}{x^2}e^x

[Matuku]

Homework Statement

Find,
$$\int \left( \frac{1}{x} - \frac{1}{x^2} \right)e^x ~dx$$

Homework Equations

None

The Attempt at a Solution

I tried integrating by parts,
$$\]\int \left( \frac{1}{x} - \frac{1}{x^2} \right)e^x ~dx\\<br /> Let ~\frac{dv}{dx}=\left( \frac{1}{x} - \frac{1}{x^2} \right), and ~u=e^x.\\<br /> \therefore v=\ln{x} + \frac{1}{x}, and ~\frac{du}{dx}=e^x\\<br /> \therefore \int \left( \frac{1}{x} - \frac{1}{x^2} \right)e^x ~dx<br /> = e^x(\ln{x} + \frac{1}{x}) - \int{ e^x(\ln{x} + \frac{1}{x})}~dx\[$$

But I can't see what to do now; the next integral is even messier than the first!

 

[MathematicalPhysicist]

I think that it won't work here, you need to develop e^x power series expansion.

 

[Citan Uzuki]

First break the integral apart thus:

$$\int \left( \frac{1}{x} - \frac{1}{x^2} \right) e^x\ \text{d}x = \int \frac{1}{x} e^x \ \text{d}x - \int \frac{1}{x^2} e^x \ \text{d}x$$

Now, integrate the first of the two resulting integrals by parts, letting $u=\frac{1}{x}$ and $\text{d}v=e^x\ \text{d}x$. You should get a very convenient cancellation.