# Definite integral involving $\sqrt{\log}$

In some old lecture notes I found the following problem: $$\int_1^e \frac{1+2x \sqrt{\log x}}{2 x \sqrt{\log x}(x+\sqrt{\log x})} \;dx$$ I just don't seem to get a handle on this rather difficult looking integral. Does anybody have some insights?

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## 1 Answer

First note that \begin{align*} \left(\sqrt{\log{x}}\right)'=\frac{1}{2x\sqrt{\log{x}}} \end{align*} Then for the integrand it follows that \begin{align*} \frac{1+2x\sqrt{\log x}}{2x\sqrt{\log x}(x+\sqrt{\log x})} &=\frac{\frac{1}{2x\sqrt{\log x}}+1}{x+\sqrt{\log x}}\\ &=\frac{(x+\sqrt{\log x})'}{x+\sqrt{\log x}}. \end{align*} This enables us to calculate \begin{align*} \int\frac{(x+\sqrt{\log x})'}{x+\sqrt{\log x}}dx=\log\left(x+\sqrt{\log x}\right) \end{align*} and therefore \begin{align*} \int^e_1\frac{(x+\sqrt{\log x})'}{x+\sqrt{\log x}}dx =\log\left(e+\sqrt{\log e}\right)-\log\left(1+\sqrt{\log 1}\right)=\log\left(e+1\right). \end{align*}

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+1: thanks, I should have seen this rather obvious substitution... – Fabian Jan 5 '13 at 12:47