Big $\theta$ of $x^2ln(x) + xln(x^2+1)$

Question

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:

$x^2ln(x) + xln(x^2+1) \le x^2ln(x) + xln(x^2+1)$
$\le x^2ln(x(x^2+1)) \le x^2ln(x^3+x)$

score 1 · Accepted Answer · answered Oct 03 '15 at 19:18

1

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 $x \ge 3$:

$$x^2\ln(x) + x\ln(x^2+1) \le x^2\ln(x) + x\ln(x^x) = x^2\ln(x)+ x^2\ln(x) = 2 x^2\ln(x)$$

answered Oct 03 '15 at 19:18

Henry

157,058

Why would it be $x \ge 3$ ? – Paradox Oct 03 '15 at 19:20
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@Paradox: because $2^2+1 \gt 2^2$ but $3^2+1 \lt 3^3$ – Henry Oct 03 '15 at 19:21

Big $\theta$ of $x^2ln(x) + xln(x^2+1)$

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