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$$\int \sqrt\frac{x^3-3}{x^{11}}\,dx$$

The form the answer takes suggests a very quick substitution should be possible. I cannot see how to obtain it in only a few steps and would be grateful for any help.

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    Hint: $\sqrt{\frac{1}{x^{11}}}=\frac{1}{x^4} \sqrt{\frac{1}{x^3}}$. What substitution you're gonna use after you divide $x^3-3$ by $x^3$ inside the square root? – Zacky Aug 11 '19 at 10:50

2 Answers2

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$$\int \sqrt\frac{x^3-3}{x^{11}}\,dx=\int\frac1{x^4}\sqrt{\frac{x^3-3}{x^3}}\,dx=\int\frac1{x^4}\sqrt{1-\frac3{x^3}}\,dx$$ The substitution made is $u=\frac1{x^3},\frac{du}{dx}=-\frac3{x^4}$: $$=-\frac13\int\sqrt{1-3u}\,du=-\frac13\left(-\frac29(1-3u)^{3/2}\right)+K$$ $$=\frac2{27}(1-3/x^3)^{3/2}+K$$

Parcly Taxel
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You could try the Ansatz $$\int x^{-11/2}(x^3-3)^{1/2}dx=ax^b(x^3-3)^c+K,$$so $$x^{-11/2}(x^3-3)^{1/2}=ax^{b-1}(x^3-3)^{c-1}\left((b+3c)x^3-3b\right),$$which has solution $a=\frac{2}{27},\,b=-\frac92,\,c=\frac32$.

J.G.
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