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How to integrate $$\frac{1}{x\sqrt{x}}$$ I don't see how could I use u substitution or integration by parts. I tried both, but it just got worse(more complex). I haven't integrate in years and I just can't warp my head around this.

Edit: Thank you everybody who helped me. It was really simple and obvious. Now it is easy.

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    $\begingroup$ $1/(x\sqrt x)=x^{-3/2}$. $\endgroup$ Commented Mar 18, 2016 at 21:28
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    $\begingroup$ Rewrite it as $x^{-3/2}$. Then use the power rule. $\endgroup$ Commented Mar 18, 2016 at 21:29
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    $\begingroup$ Please read this tutorial on how to format mathematics on this site. $\endgroup$ Commented Mar 18, 2016 at 21:36
  • $\begingroup$ Which one should I accept as an answer? I guess the first in chronological order. $\endgroup$ Commented Mar 18, 2016 at 21:38
  • $\begingroup$ Suppose that you wanted to use a substitution (not the easiest way, but for academic interest...). A substitution like $u^2=x, 2u\text du = \text dx$ is one way, but maybe more interesting would be $v=\frac 1{\sqrt x},\text dv=-\frac{\text dx}{2x\sqrt x}$: $$\int{1\over x\sqrt x}\text dx=-2\int \text dv=-2v+C=-\frac 2{\sqrt x}+C$$ $\endgroup$ Commented Mar 18, 2016 at 21:41

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Hint: $$ \frac{1}{x\sqrt{x}}=\frac{1}{x^{3/2}}=x^{-3/2} $$

now can you find a primitive?

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$$\int\frac{dx}{x\sqrt x}=\int\frac{dx}{x^{3/2}}=\int x^{-3/2}dx=\color{red}{-\frac{2}{\sqrt x}+\mathcal C}$$

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Hint: $1/(x\sqrt x) = x^{-3/2}$ so its primitive is $-2x^{-1/2}$.

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    $\begingroup$ That is more than a hint. $\endgroup$ Commented Mar 18, 2016 at 21:32
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Notice, when $n>1$:

$$\int\frac{1}{x^n}\space\text{d}x=\int x^{-n}\space\text{d}x=\frac{x^{1-n}}{1-n}+\text{C}$$

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  • $\begingroup$ Yes. I've used this. Thanks. I just overlooked the fact that I can move x inside the √ as x² $\endgroup$ Commented Mar 18, 2016 at 21:39
  • $\begingroup$ @SzőkeSzabolcs You're welcome! $\endgroup$ Commented Mar 18, 2016 at 21:40
  • $\begingroup$ Should be OK over any interval with $x > 0$, but the final expression won't be valid if $n=1$. $\endgroup$ Commented Mar 18, 2016 at 21:41
  • $\begingroup$ @Bungo Yes that's true, thats why I choose to put in $n>1$ because the antiderivative of $\frac{1}{x}$ is $\ln|x|+\text{C}$ $\endgroup$ Commented Mar 18, 2016 at 21:42

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