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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.

1 Answers 1

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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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    @André Nicolas: Thank you :) Now it's all clear.2012-11-23