$$
\int \left(\sqrt{2\log x}+ \frac{1}{\sqrt{2\log x}} \right) dx
$$
I am stuck on this problem from This years integration bee. I have tried substitution but it is not giving the correct answer which is $x\sqrt{2\log x}$
I supposed $\sqrt{2\log x}$ as $t$ and differentiated it wrt $x$, and substituted it in the above integral. But the solution has an extra $(\frac {2\log x+1}{3})$, i dont know how and why?
Best Answer
Note that: $$\frac{\mathrm{d}}{\mathrm{d}x} \sqrt{2 \log x}=\frac{1}{x\sqrt{2 \log x}}$$ Using the product rule, you have: $$ \int \left(\sqrt{2\log x}+ \frac{1}{\sqrt{2\log x}} \right) \mathrm{d}x= \int \left(\sqrt{2\log x}+ \frac{x}{x\sqrt{2\log x}} \right) \mathrm{d}x=\int \left(\frac{\mathrm{d}}{\mathrm{d}x} x\sqrt{2\log x}\right) \mathrm{d}x=x\sqrt{2\log x}+k $$