Regular Season Problem 11 from 2023 MIT Integration Bee

Question

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

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?

Answer 1

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

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

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

Answer 2

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: $$\begin{align} \int \left(\sqrt{2\log x} + \frac{1}{\sqrt{2 \log x}}\right)\ dx& = \int\left( t + \frac{1}{t}\right)t e^{t^2/2} \ dt\\& = \int e^{t^2/2}(t^2 + 1)\ dt\\& =\int t^2 e^{t^2/2} dt + \color{blue}{ \int e^{t^2/2} \ dt}\\&\overset{\tt {IBP}} =\int t^2 e^{t^2/2} dt\color{blue}{ + t e^{t^2/2}- \int t^2 e^{t^2/2} dt} + C\\& = t e^{t^2/2} + C\\& = x\sqrt{2\log(x)} + C\end{align}$$

Answer 3

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

Answer 4

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