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thomas5267

  • 3 years ago

Rewrite \(\int^3_2\frac{3x}{x^2-4x+5}\,dx\) into the form of \(\int^b_a\frac{mt+c}{t^2+k^2}\) or \(\int^b_a\frac{mt+c}{t^2-k^2}\). Completing the square, as far as I can see, wouldn't work this time. What should I do with this?

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  1. Callisto
    • 3 years ago
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    It works :)

  2. Callisto
    • 3 years ago
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    \[\int^3_2\frac{3x}{x^2-4x+5}dx=\int^3_2\frac{3x}{x^2-4x+(4+1)}dx=\int^3_2\frac{3x}{(x-2)^2+1}dx\]In this case, t=x-2, k=1

  3. thomas5267
    • 3 years ago
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    *facepalm* I am such an idiot.

  4. Callisto
    • 3 years ago
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    No, you're not..

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