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.
Best Answer
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.