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$$\int \left\{ \frac{1}{\sqrt[3]{x}+\sqrt[4]{x}}+\frac{\log(1+\sqrt[6]{x})}{\sqrt[3]{x}+\sqrt[2]{x}} \right\}dx$$ Found this question on a worksheet. The first part seems elementary but I can't evaluate it using any standard formulas. I haven't been taught Integration by parts yet, is this question meant to be evaluated using it ? And for the second part, I tried substitution but gave no results. Even just an idea(or strategy) of how to solve is enough, since I have plenty of time left for me to solve it.

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    $\begingroup$ See math.stackexchange.com/questions/529961/… $\endgroup$ Sep 21 '16 at 16:25
  • $\begingroup$ @labbhattacharjee thanks! $\endgroup$
    – H G Sur
    Sep 21 '16 at 16:34
  • $\begingroup$ the solution of this integral looks terrible $\endgroup$ Sep 21 '16 at 16:40
  • $\begingroup$ @Dr.SonnhardGraubner May I know your answer so that I can compare with mine when I solve it ? $\endgroup$
    – H G Sur
    Sep 21 '16 at 16:40
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Usually with such integrands, we'd make the substitution $u =\sqrt[12]{x}$, so that $du = {1\over12}x^{-11/12}\; dx = {1 \over 12u^{11}}\; dx$. Your integral becomes $$\int \left(\frac{1}{u^4+u^3} + \frac{\log(1+u^2)}{u^4+u^6}\right) 12u^{11} \; du.$$ Still a bit icky.

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This answer is a continuation of that by B.Goddard. His integral has many rational terms which may be integrated by standard method, it also contains terms $$\int u^5\ln(1+u^2)du$$ which can also be integrated by parts. Finally we arrive at $$\int \frac{u}{1+u^2}\ln(1+u^2)dx=\frac{1}{4}(\ln(1+u^2))^2$$

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you can compare your solution with this here $$\frac{3 x^{2/3}}{2}-\frac{12 x^{7/12}}{7}-\frac{12 x^{5/12}}{5}+\frac{4 \sqrt{x}}{3}+\frac{11 \sqrt[3]{x}}{2}-4 \sqrt[4]{x}-5 \sqrt[6]{x}-12 \sqrt[12]{x}-3 \log ^2\left(\sqrt[6]{x}+1\right)+\left(2 \sqrt[3]{x}-3 \sqrt[6]{x}+6\right) \sqrt[6]{x} \log \left(\sqrt[6]{x}+1\right)+11 \log \left(\sqrt[6]{x}+1\right)+12 \log \left(\sqrt[12]{x}+1\right)$$

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