# Integral problem from Hardy book.

• June 20th 2010, 09:34 AM
Also sprach Zarathustra
Integral problem from Hardy book.
Can someone solve me this problem...
• June 20th 2010, 10:27 AM
p0oint
Hint: $ax^2+2bx+c=a(x^2+\frac{2b}{a}x+\frac{c}{a})=a((x+\ frac{b}{a})^2-\frac{b^2}{a^2}+\frac{c}{a})$
$a((x+\frac{b}{a})^2+\frac{ac-b^2}{a^2})=a((x+\frac{b}{a})^2+(\frac{\sqrt{ac-b^2}}{a})^2)=a(\frac{\sqrt{ac-b^2}}{a})^2(\frac{(x+\frac{b}{a})^2}{(\frac{\sqrt{ ac-b^2}}{a})^2}+1)$

Now just make a proper substitution $u = \frac{(x+\frac{b}{a})}{(\frac{\sqrt{ac-b^2}}{a})}$
• June 20th 2010, 10:36 AM
mfetch22
This might come in hadny considering this is an integral:

$tan^{-1}(x) = arctan(x) = \int_0^x\frac{1}{z^2+1}dz$

I'm not exactly sure how to solve this, but if you see an expression of the form $\frac{1}{x^2+1}$, then be on the look out for simplification by the arctangent integral indentity
• June 20th 2010, 10:44 AM
p0oint
@mfetch22 you're right, check out my last post.