# Regular Season Problem 11 from 2023 MIT Integration Bee

$$\int \left(\sqrt{2\log x}+ \frac{1}{\sqrt{2\log x}} \right) dx$$

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?

• Welcome to Math.SE! It is usually better to format with MathJaX, which I did for you here. In addition, please update the question with what you have tried, and we will then be able to guide you properly. Commented Aug 3, 2023 at 12:46
• $(fg)'=f'g+fg'$ Commented Aug 3, 2023 at 12:47
• Split the integral into the sum of the two integrals and do integration by parts on $\int 1\cdot \sqrt{2\log(x)}$. The $1$ to be integrated, the $\sqrt{2\log(x)}$ to be differentiated.
– NDB
Commented Aug 3, 2023 at 12:52
• $I= x\sqrt{2\ln x}$
– Ace
Commented Aug 3, 2023 at 12:55

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: $$\int \left(\sqrt{2\log x}+ \frac{1}{\sqrt{2\log x}} \right) \mathrm{d}x= \int \left(\sqrt{2\log x}+ \frac{x}{x\sqrt{2\log x}} \right) \mathrm{d}x=\int \left(\frac{\mathrm{d}}{\mathrm{d}x} x\sqrt{2\log x}\right) \mathrm{d}x=x\sqrt{2\log x}+k$$

Using the substitution$$\sqrt{2 \log(x)} = t$$, we have \boxed{\begin{aligned} 2\log(x) &= t^2\\\log(x) & = \frac{t^2}{2}\\x &= e^{t^2/2}\\dx &= t e^{t^2/2} \ dt \end{aligned}}

With this the given integral can be evaluated as follows: \begin{align} \int \left(\sqrt{2\log x} + \frac{1}{\sqrt{2 \log x}}\right)\ dx& = \int\left( t + \frac{1}{t}\right)t e^{t^2/2} \ dt\\& = \int e^{t^2/2}(t^2 + 1)\ dt\\& =\int t^2 e^{t^2/2} dt + \color{blue}{ \int e^{t^2/2} \ dt}\\&\overset{\tt {IBP}} =\int t^2 e^{t^2/2} dt\color{blue}{ + t e^{t^2/2}- \int t^2 e^{t^2/2} dt} + C\\& = t e^{t^2/2} + C\\& = x\sqrt{2\log(x)} + C\end{align}

• yes it took me another hour to understand where did you apply the formula, and the moment i figured it out you commented😂😂 Commented Aug 3, 2023 at 17:29
• thanks for the help Commented Aug 3, 2023 at 17:30

Integration by parts helps us obtain \begin{aligned} \int \frac{1}{\sqrt{2 \log x}} d x & =\sqrt{2} \int x d(\sqrt{\log x}) \\ & =\sqrt{2} x \sqrt{\log x}-\int\sqrt{2 \log x} d x \end{aligned} Rearranging yields the result $$\int\left(\sqrt{2 \log x}+\frac{1}{\sqrt{2 \log x}}\right) d x=x \sqrt{2 \log x}+C$$

• We can also do IBP with the other term. Commented Dec 4, 2023 at 11:28
• Yes, we can. Thank you.
– Lai
Commented Dec 5, 2023 at 2:54

Integration by parts give an instant solution like a test question.

We can also make the substitution $$x=e^t$$:

$$\int\left(\sqrt{2 \log x}+\frac{1}{\sqrt{2 \log x}}\right) d x =\int\left(\color{red}{\sqrt2\sqrt t }\color{green}{ e^t}+\color{red}{\frac1{\sqrt2\sqrt t}}\color{green}{e^t}\right)dt \stackrel{u=\sqrt{2t}, v=e^t}{=}\int(uv'+u'v)dt=\int(uv)'dt=uv+C=e^t\sqrt{2t}+C =x \sqrt{2 \log x}+C$$