1. ## Definite Integral

$\displaystyle \int_{0}^{\infty}\frac{\ln(x)}{x^2+2x+4}dx$

3. ## Re: Definite Integral

Have you tried integration by parts?

Of course, romsek's solution works if all you need is the answer - you will have to put the correct polynomial in the denominator.

- Hollywood

4. ## Re: Definite Integral

My Trial solution:: $\int_{0}^{\infty}\frac{\ln(x)}{x^2+2x+4}dx = \int_{0}^{\infty}\frac{\ln(x)}{(x+1)^2+\left(\sqrt {3}\right)^2}dx$

Now Let $(x+1) = \sqrt{3}\tan \theta,$ Then $dx = \sqrt{3}\sec^2 \theta d\theta$

So Integral convert into $\displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\frac{\ln\left (\sqrt{3}\tan \theta - 1\right)}{3\sec^2 \theta}\cdot \sqrt{3}\sec^2 \theta d\theta = \frac{1}{\sqrt{3}}\int_{\frac{\pi}{6}}^{\frac{\pi} {2}}\ln\left(\sqrt{3}\tan \theta - 1\right)d\theta$

$\displaystyle = \frac{1}{\sqrt{3}}\cdot \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\ln\left(\frac {\sqrt{3}\sin \theta - \cos \theta}{\cos \theta}\right)d\theta$

$\displaystyle = \frac{1}{\sqrt{3}}\int_{\frac{\pi}{6}}^{\frac{\pi} {3}}\ln \left(\sqrt{3}\sin \theta -\cos \theta\right)d\theta - \frac{1}{\sqrt{3}}\int_{\frac{\pi}{6}}^{\frac{\pi} {3}}\ln (\cos \theta)d\theta$

Now How can I solve after that...

Help me

Thanks