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$$ \int \frac{\sqrt[3]{x}+1}{\sqrt[4]{x^3}(\sqrt{x} - \sqrt[6]{x})dx} $$

So I used substitution method: $$ x = t^{12} $$ $$ dx = 12t^{11}dt $$ and I ended up with a very weird integral... could you help me?

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3  
Maybe it is better to put $dx$ in the numerator? –  Boris Novikov Mar 12 '13 at 12:36
1  
What weird integral? –  Ron Gordon Mar 12 '13 at 12:37
    
The problem is, that dx is in the denominator and I think I can't touch it. –  TomDavies92 Mar 12 '13 at 12:44
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With $dx$ in the denominator, the rest of us do not know what that notation means, so you will have to supply a definition. –  GEdgar Mar 12 '13 at 13:04
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@GEdgar reminds me of an old joke. :) I moved the $dx$ out of the denominator. –  Thomas Andrews Mar 12 '13 at 13:04
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1 Answer

up vote 6 down vote accepted

Actually, this is beautiful. When you substitute $x=t^{12}$, you get

$$12 \int dt \frac{t^4-1}{t^4+1}$$

Use partial fractions to show that

$$\frac{t^4-1}{t^4+1} = 1 + \frac{1/2}{t+1} - \frac{1/2}{t-1} + \frac{1}{1+t^2}$$

Then the integral is

$$12 t + 6 \log{\left (\frac{t+1}{t-1}\right)} + 12 \arctan{t} + C$$

where $C$ is an integration constant. Substitute back $t=x^{1/12}$ and you are done.

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Thank you !:) Now I see it. It seems you just moved dx in the numerator and then substituted. Is it correct? –  TomDavies92 Mar 12 '13 at 12:52
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Yes. The $dx$ in the denominator is nonsensical. –  Ron Gordon Mar 12 '13 at 12:52
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