It surely should be solved using integration by parts, but I don't know how to proceed: $$ \int\frac{1}{x\sqrt{1-(\ln x)^2}}dx $$ I tried having $du=1/x$ and thus $u=\ln x$, but then I don't know how to deal with the $(\ln x)^2$ in the denominator. Can anybody help please?
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1Substitution, not parts.Let $u=\ln x$. Then we are finding $\int \frac{1}{(1-u^2)^{1/2}},du$, probably familiar. – André Nicolas Jan 13 '15 at 12:29
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Following the suggestion in Andre Nicolas's comment, you can let $u=\ln x, du=\frac{1}{x}dx$ to get
$\;\;\;\;\displaystyle\int\frac{1}{\sqrt{1-u^2}}\;du = \sin^{-1}u+C=\sin^{-1}(\ln x)+C$.
user84413
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Substitute $$\ln x=\sin y.$$ $$\int\frac{1}{x\sqrt{1-(\ln x)^2}}dx=\int dy=y+c$$ Therfore$$\int\frac{1}{x\sqrt{1-(\ln x)^2}}dx=\arcsin(\ln x)+c,$$ where $c$ is an arbitary constant.
Bumblebee
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4@user84413: It seems like $$\begin{align}\int\frac1{x\sqrt{1-\log(x)^2}}\mathrm{d}x &=\int\frac1{\sqrt{1-\log(x)^2}}\mathrm{d}\log(x)\ &=\int\frac1{\sqrt{1-\sin^2(y)}}\mathrm{d\sin(y)}\ &=\int\mathrm{d}y\end{align}$$ – robjohn Jan 13 '15 at 18:25
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