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I've reached a pretty weird integral $$\int_0^{5} \frac{\ln(y)}{(y+3)\sqrt{y}} dy,$$ And I'm having some difficulties starting from the $u$-substitution method. I had the intuition that I may take $\sqrt{y} = u$ and thus $\frac{1}{2\sqrt{y}}dy = du.$ However, this method seems to get tangled with the issues related to the natural log in the numerator. I felt that I could start on integration by parts, but then I thought that there may be a cleaner method with partial fractions. Could someone give me some suggestions on either method in this problem?

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    $\begingroup$ There isn't any clean solution. This cannot be expressed in terms of elementary functions. As far as the definite integral goes, the function is not symmetrical about its limits or any such thing, so I guess evaluating it using numerical methods is the only way to solve it. $\endgroup$ – Shailesh Dec 29 '15 at 2:17
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    $\begingroup$ wolfram alpha can't find any closed form for this integral, so I suspect this to be pretty hard(if possible...) $\endgroup$ – math635 Dec 29 '15 at 2:18
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    $\begingroup$ Apparently you can express the result using polylogarithms: wolframalpha. $\endgroup$ – anderstood Dec 29 '15 at 2:22
  • $\begingroup$ You say you've "reached" a pretty weird integral - does that mean that your attempted solution of a problem led (after some steps) to this integral? In which case, please post the original problem, because there could have been a mistake in your working that led to this difficult integral. $\endgroup$ – Deepak Dec 29 '15 at 2:24
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    $\begingroup$ Substitution and parts reduces to $\int\frac{\arctan x}{x}dx$ I think this is a well known non integrable in elementary functions. $\endgroup$ – Rene Schipperus Dec 29 '15 at 2:25
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$$\int \frac{\log(y)}{(y+3)\sqrt{y}} dy$$ $t=\sqrt{y},\;\; y=t^2,\;\;dy=2t\,dt$ $$=4\int \frac{\log(t)}{t^2+3} dt$$ $u=\log(t),\;\;du=\frac 1t,\;\; v=\frac{1}{\sqrt{3}}\arctan\left(\frac{t}{\sqrt{3}}\right),\;\; dv = \frac{1}{t^2+3}$ $$=4\left(\frac{\log(t)}{\sqrt{3}}\arctan\left(\frac{t}{\sqrt{3}}\right)-\frac{1}{\sqrt{3}}\int\frac{\arctan\left(\frac{t}{\sqrt{3}}\right)}{t}dt\right)$$ Looking at the final integral, we get that $$\int\frac{\arctan\left(\frac{t}{\sqrt{3}}\right)}{t}dt$$ $$=\frac{i}{2}\left(\int\frac{\log\left(1-\frac{t}{\sqrt{3}}\right)}{t}dt-\int\frac{\log\left(1+\frac{t}{\sqrt{3}}\right)}{t}dt\right)$$ $$=\frac{i}{2}\left(I_1 - I_2\right)$$ $u=\frac{t}{\sqrt{3}},\;\; t=u\sqrt{3},\;\; dt=du\sqrt{3}$ $$I_1 = -\int\frac{-\log(1-u)}{u}du = -\operatorname{Li_2}(u) = -\operatorname{Li_2}\left(\frac{t}{\sqrt{3}}\right)$$ $u=\frac{-t}{\sqrt{3}},\;\; t=-u\sqrt{3},\;\; dt=-du\sqrt{3}$ $$I_2 = -\int\frac{-\log(1-u)}{u}du = -\operatorname{Li_2}(u) = -\operatorname{Li_2}\left(\frac{-t}{\sqrt{3}}\right)$$ Putting this all together, we get $$\frac{4\log(t)}{\sqrt{3}}\arctan\left(\frac{t}{\sqrt{3}}\right)+\frac{i}{2\sqrt{3}}\left[\operatorname{Li_2}\left(\frac{t}{\sqrt{3}}\right)+\operatorname{Li_2}\left(\frac{-t}{\sqrt{3}}\right)\right]$$ $$=\color{red}{\frac{4\log(t)}{\sqrt{3}}\arctan\left(\frac{t}{\sqrt{3}}\right)+\frac{i}{4\sqrt{3}}\operatorname{Li_2}\left(\frac{t^2}{3}\right)}$$ This uses the Polylogarithm. Here is a link to the Wikipedia page on the topic.

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    $\begingroup$ Use $\text{Li}_2(a)+\text{Li}_2(-a)=\dfrac{\text{Li}_2\big(a^2\big)}2$ to simplify the final result. $\endgroup$ – Lucian Dec 29 '15 at 5:47
  • $\begingroup$ @Lucian Good catch! Completely missed that! $\endgroup$ – Brevan Ellefsen Dec 29 '15 at 6:12
  • $\begingroup$ Classy stuff @BrevanEllefsen $\endgroup$ – Kugelblitz Dec 29 '15 at 7:47
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    $\begingroup$ I'm sorry if I'm missing something obvious but I don't quite understand how you got from the arctan integral to the log integrals :-) could you clarify? $\endgroup$ – Ant Dec 29 '15 at 10:29
  • $\begingroup$ @Ant He's using the complex logarithmic forms of the $\arctan$ function. See wikipedia $\endgroup$ – GregRos Dec 29 '15 at 14:26
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This is more of a comment since I'm not sure this is what you're looking for but if you use $t=\sqrt y$ then $dy=2t\ dt$. The integrand becomes

$$2t \frac{\ln t^2}{(t^3+3t)}=4\frac{\ln t}{t^2+3}$$

Then by parts, $u=4\ln t$ and $dv=\frac{1}{t^2+3}$. This gives $du=\frac{4}{t}$ and $dv$ is a function of arctan.

This leave you with a new integrand which looks like $\frac{\arctan t}{t}$ which is POSSIBLE with polylogarithmic functions but not with elementary functions.

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