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Find $$\int \frac{\mathrm dx}{\sqrt{x}+\sqrt[3]{x}}$$

I substituted $t = \sqrt x$ so $x = t^2$ and $\mathrm dx = 2t \mathrm dt$. I got to the

$$ 2\int \frac{dt}{1+t^{-\frac13}} $$

I'm not sure, if that is right. I still do lots of mistakes, but even if it would be right, I don't know, what to do next.

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1 Answer 1

up vote 9 down vote accepted

Hint: Instead of the substitution you used, start by letting $x=t^6$. The problem collapses.

Remark: Your calculation was correct. After that, if you put $t=u^3$, the calculation can be completed.

The initial substitution $x=t^6$ is "natural." It lets us get rid of all roots. Trigonometric substitutions, and hyperbolic function substitutions, have a similar motivation.

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3  
@user50222: Note that whenever you have an integral as $$\int R(x,x^{\frac{p_1}{q_1}},x^{\frac{p_2}{q_2}},...,x^{\frac{p_k}{q_k}})dx$$, then one good substitution will be $x=t^{\text{lcm}(q_1,q_2,...,q_k)}$ –  Babak S. Nov 23 '12 at 18:17
    
Thank you. That's a good idea. –  user50222 Nov 23 '12 at 19:06
    
But I'm still not able to complete the calculation. Now I got 6* integral of t^3/(t+1)dt. Maybe I 'm just tired, but.. I don't know, what to do next... –  user50222 Nov 23 '12 at 19:11
1  
One of two things: 1) Divide the polynomial $t^3$ by $t+1$. We get $t^2-t+1-\frac{1}{t+1}$. Now easy! Or else 2) Let $u=t+1$. Then top becomes $u^3-3u^2+3u-1$, bottom is $u$, divide, integrate. –  André Nicolas Nov 23 '12 at 19:16
    
@André Nicolas: Thank you :) Now it's all clear. –  user50222 Nov 23 '12 at 19:30

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