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$$\int {\frac{1}{x^\frac13+x^\frac14}+\frac{log(1+x^\frac16)}{x^\frac13+x^\frac12}}dx$$ I have to sovle this function integral. Please help me out I have tried a lot but its approach is very long. I have also seen its solution bit still its quite long approach. Please suggest the shortest possible method.

Note: The solution should also be in elementary form.

Adesh Tamrakar
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    Are you sure you've copied it accurately? integrals.wolfram.com gives something very ugly and non-elementary. – Anton Sherwood Sep 30 '15 at 07:57
  • Yes that's the question. I have copied it correctly and it's only elementary function. – Adesh Tamrakar Sep 30 '15 at 08:01
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    Four halves and six halves? Why not just write $2$ and $3$? – Mike Sep 30 '15 at 08:07
  • The result contains non-elementary functions (the first term does not, but the second does). Where did you find this problem? – mickep Sep 30 '15 at 08:40
  • I have picked up this problem from elementary function integration topic so don't worry the question is in elementary functions . – Adesh Tamrakar Sep 30 '15 at 08:44
  • Nope, the second part is non-elementary. It will typically involve the dilogarithm function $\text{Li}_2$. – mickep Sep 30 '15 at 08:54
  • Error 504: Question missing context. – Aditya Agarwal Sep 30 '15 at 08:54
  • I hope you won't get in trouble if not being able to find a primitive in terms of elementary functions... – mickep Sep 30 '15 at 09:07
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    The question is picked from A. Das Gupta Problems Plus in IIT Mathematics. You can just Google the book if you have a doubt. I also have the solution but its very long and is elementary form and I just want a shortest possible method. – Adesh Tamrakar Sep 30 '15 at 09:09
  • Please post the final answer so that it is proved I'm wrong. – mickep Sep 30 '15 at 09:09
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    $$12\left[\frac{z^8}{8} -\frac{z^7}{7} +\frac{z^6}{6} -\frac{z^5}{5} +\frac{z^4}{4} -\frac{z^3}{3} +\frac{z^2}{2}-z\right] +12\log(z+1) +6\left[\frac13\times\alpha\times e^\alpha -\frac19\times e^\alpha -\frac32\times e^\alpha2 +\frac34\times e^2\alpha +3\alpha\times e^\alpha -3e^\alpha -\frac{\alpha^2}{2}\right] +\text{constant}$$ – Adesh Tamrakar Sep 30 '15 at 09:22
  • Where $z=x^{\frac{1}{12}}$ and $\alpha = \log(1+x^3)$. – Adesh Tamrakar Sep 30 '15 at 09:27
  • I have now entered every possible form of that expression that I can guess, and differentiated, and the result has not been close to the original function. Please update your question with the correct expression. Be careful when you write. – mickep Sep 30 '15 at 09:46
  • Let us continue this discussion in chat. – Adesh Tamrakar Sep 30 '15 at 12:36
  • OK, so now the question has changed considerably. It now has an elementary primitive, indeed. – mickep Sep 30 '15 at 12:57

1 Answers1

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For the first part, write $$ \frac{1}{x^{1/4}+x^{1/3}}=\frac{1}{(1+x^{1/12})x^{1/4}} $$ and then let $u=x^{1/12}$. Using a geometric sum, you end up with $$ 12\int \frac{u^8}{1+u}\,du=12\int -1+u-u^2+u^3-u^4+u^5-u^6+u^7+\frac{1}{1+u}\,du. $$ which is easy to integrate. I leave it to you.

For the second part, write $$ \frac{\log(1+x^{1/6})}{x^{1/3}+x^{1/2}}=\frac{\log(1+x^{1/6})}{(1+x^{1/6})x^{1/3}}, $$ and then let $u=x^{1/6}$. You end up with $$ 6\int\frac{u^3\log(1+u)}{1+u}\,du. $$ Let $s=\log(1+u)$, and you will find (again, modulo constant), $$ 6\int (e^s-1)^3s\,ds. $$ This is easily calculated, integrating by parts.

mickep
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