I am working on finding the big $\theta$ of $x^2ln(x) + xln(x^2+1)$ but have never done it before with logarithms. How would I handle this problem? I also know that n is big $\Omega$ of nln(n) which I feel has something to do with this question... This is what I have tried for big Oh but although it seems incorrect:
$x^2ln(x) + xln(x^2+1) \le x^2ln(x) + xln(x^2+1)$
$\le x^2ln(x(x^2+1)) \le x^2ln(x^3+x)$
Your final expression for $x \ge 2$ is less than or equal to $x^2\ln(x^4)= 4 x^2\ln(x)$.
Alternatively you could say for $x \ge 3$:
$$x^2\ln(x) + x\ln(x^2+1) \le x^2\ln(x) + x\ln(x^x) = x^2\ln(x)+ x^2\ln(x) = 2 x^2\ln(x)$$