Use "partial fractions" to write as . Then integrate (you will get logarithms).

What happens as m-->0+

Compute integral (1 to 2) dx/((x^2)+n) if n is a small positive number.

Now that looks like an "arctan" problem to me. . If you let then . Of course, . When x= 1, u= and when x= 2, u= so your integral becomes [tex]\int_{1/\sqrt{n}}^{2/\sqrt{n}}\frac{\sqrt{n}du}{nu^2+ n}= 1/\sqrt{n}int_{1/\sqrt{n}}^{2/\sqrt{n}}\frac{du}{u^2+ 1}

What happens when n-->0+

From the look of it, these integrals will be equal as both both n and m approach 0...the problem is I don't know how to show this.....