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**1. Homework Statement**

[tex]

\int{\frac{e^x}{x^2}dx}

[/tex]

**2. Homework Equations**

- Integration by substitution
- Integration by parts: [itex]\int{u\ dv}=uv\ -\ \int{v\ du}[/itex]

**3. The Attempt at a Solution**

Since it was clear that integration by substitution would not work, I tried integration by parts. Since the [itex]e^x[/itex] term would not be affected whatsoever with the application of differentiation or integration, I worked out the [itex]x^2[/itex] term instead. So, I took [itex]u=e^x[/itex] and [itex]dv=\frac{1}{x^2}\ dx[/itex]. It resulted in:

[tex]

\int{\frac{e^x}{x^2}dx}=-\frac{e^x}{x}+\int {\frac{e^x}{x}dx}

[/tex]

For the last term, I took [itex]u=e^x[/itex] and [itex]dv=\frac{1}{x}\ dx[/itex]. It resulted in:

[tex]

\int{\frac{e^x}{x^2}dx}=-\frac{e^x}{x}+e^x\ ln(x)-\int{(ln x)(e^x)dx}

[/tex]

Beyond this point, if I took [itex]u=ln(x)[/itex] and [itex]dv=e^x\ dx[/itex], I would just undo the previous steps. If I took [itex]u=e^x[/itex] and [itex]dv=ln(x)\ dx[/itex], it resulted in:

[tex]

\int{\frac{e^x}{x^2}dx}=-\frac{e^x}{x}\ +\ e^x\ ln(x)\ -\ e^x\left ( x\ ln(x)\ -\ x \right )\ +\ e^x\ -\ x\ e^x\ +\ \int{x\ ln(x)\ e^x\ dx}}

[/tex]

The last term certainly shows that this technique won't solve the problem at hand because it will continue forever.

How should I attack this problem?

Thank you.

Best regards,

Eus