$$ \int \frac{\sqrt[3]{x}+1}{\sqrt[4]{x^3}(\sqrt{x} - \sqrt[6]{x})dx} $$
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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3Maybe it is better to put $dx$ in the numerator? – Boris Novikov Mar 12 '13 at 12:36
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1What weird integral? – Ron Gordon Mar 12 '13 at 12:37
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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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2With $dx$ 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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1@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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@TomDavies92 The integral doesn't mean anything with the $dx$ in the denominator. $dx$ is not a number that can be plugged into any formula, it is notation. If the problem really has $dx$ in the deonominator, then the problem is unanswerable because the notation is meaningless. – Thomas Andrews Mar 12 '13 at 14:07
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(But I reverted my edit since you insist that the $dx$ has to be in the denominator.) Either that was a typo in your homework, or the creator of the problem is having some fun at your expense. – Thomas Andrews Mar 12 '13 at 14:19
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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.
Ron Gordon
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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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