Can someone explain how to integrate it?
$$\int\frac{(x^2+1)dx}{x\sqrt{x^4+3x^3-2x^2-3x+1}},x>1$$ I know that I should use a substitution but I can't figure out. I should make some transformations under the root.
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Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – CrSb0001 Jun 05 '23 at 19:49
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Did you try it using wolfram alpha? – coffeemath Jun 05 '23 at 19:55
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I know that I should use a substitution but I can't figure out. I should make some transformations under the root – James Jun 05 '23 at 19:55
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Hint: the expression in the square root is a perfect square – jgd1729 Jun 05 '23 at 19:56
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@jgd1729 you meat that we have (x^2-1)(x^2+3x-1) under the root? – James Jun 05 '23 at 20:02
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@James I mean that we have $(x^2 +\frac{3}{2}x-1)^2$ under the root. – jgd1729 Jun 05 '23 at 20:05
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@James I enclosed your comment into the question so that your effort can be seen.. – user376343 Jun 05 '23 at 20:08
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No, it is not a perfect square @jgd1729 – MathFail Jun 05 '23 at 20:14
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@MathFail you are correct, I did the algebra too quickly and missed a term – jgd1729 Jun 05 '23 at 20:17
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@jgd1729 it will be coefficient under root 1/4 with x^2 not a 2 – James Jun 05 '23 at 20:17
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@coffeemath it gives me wrong answer – James Jun 05 '23 at 20:22
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@CrSb0001 I think that I need get sqrt(t^2+k) under the root on last stage. – James Jun 05 '23 at 20:29
2 Answers
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Hint:
\begin{align} &\int\frac{x^2+1}{x\sqrt{x^4+3x^3-2x^2-3x+1}}dx\\ =&\int\frac{1+\frac1{x^2}}{\sqrt{x^2+3x-2-\frac3{x}+\frac1{x^2}}}dx =\int\frac{d(x-\frac1{x})}{\sqrt{(x-\frac1x)^2+3(x-\frac1x)}}\\ \end{align}
Quanto
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$$\int\frac{x^2+1}{x\sqrt{x^4+3x^3-2x^2-3x+1}}dx\\~\\＝\int\;\frac {\frac {1}{x^2}＋1}{x\sqrt {1＋\frac {3}{x}－\frac {2}{x^2}-\frac {3}{x^3}＋\frac {1}{x^4}}}dx\\~\\＝\int\;\frac {\frac {1}{x^2}＋1}{\sqrt {x^2＋3x－2-\frac {3}{x}＋\frac {1}{x^2}}}dx\\~\\＝\int\;\frac {\frac {1}{x^2}＋1}{\sqrt {\left({x-\frac {1}{x}}\right)^2＋3\left({x-\frac {1}{x}}\right)}}dx$$
Now : let u = $x-\frac{1}{x}$
Just a start, now try to complete the solution on your own. I think you can do that.
Mostafa
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1Can you help me with another one, please? You can find question in my profile. – James Jun 06 '23 at 04:02
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