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:



Now $I_1$ becomes

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


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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.