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Evaluate the following integral: $$\int \left(\frac{1}{\sqrt[3]x +\sqrt[4]x} + \frac{\ln(1+\sqrt[6]x)}{\sqrt[3]x +\sqrt x}\right)\,dx$$


My Attempt:

Let $$I=\int \Big(\frac{1}{\sqrt[3]x +\sqrt[4]x} + \frac{\ln(1+\sqrt[6]x)}{\sqrt[3]x +x}\Big)dx$$ and $$ I_1 = \int \left(\frac{1}{\sqrt[3]x +\sqrt[4]x} \right)\,dx$$ $$ I_2= \int \left(\frac{\ln(1+\sqrt[6]x)}{\sqrt[3]x +x}\right)\,dx$$ $$\Downarrow$$ $$I=I_1 + I_2$$

Consider $I_1$:

$$\int \left(\frac{1}{\sqrt[3]x +\sqrt[4]x} \right)\,dx$$

Introduce the substitution:

$$x=t^{12}$$

$$dx=12t^{11}\,dt$$

Now $I_1$ becomes

$$\int \left(\frac{1}{t^3+t^4}\right)(12t^{11})\,dt$$

Simplyfying:

$$\int \left(t^7-t^6+t^5-t^4+t^3-t^2+t-1+\frac{t^3}{t^3+t^4}\right)dt$$

Further simplifying, we obtain:

$$I_1=\left(\frac{t^8}{8}-\frac{t^7}{7}+\frac{t^6}{6}-\frac{t^5}{5}+\frac{t^4}{4}-\frac{t^3}{3}+\frac{t^2}{2} - t+ \log(1+t) \right) + C$$


I tried proceeding in a similar manner for the second part ($I_2$), but I think I've hit a wall. Could someone explain how I can solve the second part? All forms of help are appreciated.

Additional information: This is an IIT-JEE integral.

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  • $\begingroup$ Are you sure about your computation for the first one? It appears that the final ratio should give you a log not the arctan. For the second, I also do not see how to proceed, and wolfram gives something horribly non elementary. $\endgroup$ – qbert May 6 at 15:29
  • $\begingroup$ You have copied it wrong? Can you put a photo with the exercise? $\endgroup$ – Zacky May 6 at 15:38
  • $\begingroup$ Just something to note when attempting these hard integrals: sometimes its worth thinking why they have added two terms together if both integrals are possible individually. There may be significance to the fact that there's two terms - like reversing the product rule, but where each integral isn't possible individually. (But I don't think that works in this example, since you've done one of the integrals.) $\endgroup$ – John Doe May 6 at 15:44
  • $\begingroup$ @Threesidedcoin I've cross checked, there's no mistake. $\endgroup$ – ExtremeRaider May 6 at 15:44
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    $\begingroup$ @NikilKumar For example $$\int\frac1{6x}\left[\frac1{\sqrt[6]x+2\sqrt[3]x+\sqrt x}-\frac{\ln(1+\sqrt[6]x)(2\sqrt[3]x+3\sqrt x)}{x^{\frac23}+2x^{\frac56}+x}\right]\mathrm dx$$ comes from differentiating the following by the product rule $$\frac{\ln(1+\sqrt[6]x)}{\sqrt[3]x+\sqrt x}$$Of course that was quite derived, but I've seen other more simple looking sums that can be integrated in such a way (such as in MIT Integration Bees) $\endgroup$ – John Doe May 6 at 17:54
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There's a typo from OP, (check the photo in comments) it should be: $$I_2=\int \frac{\ln(1+\sqrt[6]{x})}{\sqrt[3]{x}+\sqrt{x}}dx$$ Now it's easy. Set the common power of the roots to $t$ so $x=t^6$. $$I_2=6\int \frac{t^3\ln(1+t)}{1+t}dt=6\int \left(t^2-t+1-\frac{1}{1+t} \right)\ln(1+t)dt$$ The first three integrals require integration by parts and the last one is the square of that logarithm divided by two.

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