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thomas5267

  • 2 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 \(\int^b_a\frac{mt+c}{t^2-k^2}\). Not even a clue how to do it. Please help.

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  1. Callisto
    • 2 years ago
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    Completing square!?

  2. Callisto
    • 2 years ago
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    Consider the denominator: \[x^2+4x+5=(x^2 +4x+4-4)+5 = (x+2)^2 -4+5 =...\]

  3. thomas5267
    • 2 years ago
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    I don't think so. Completing the square will yield \((x+a)^2+b\), which is not what we want...

  4. Callisto
    • 2 years ago
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    Hmm... \[\int^3_2\frac{3x}{x^2+4x+5} dx \]\[= \int^3_2\frac{3x}{x^2+4x+4-4+5} dx\]\[=\int^3_2\frac{3x}{(x+2)^2-4+5} dx\]\[=\int^3_2\frac{3x}{(x+2)^2+1} dx\]So, in this case, t= x+2, k=1 For the numerator, we want mt+c, and we have 3x. m must be 3. 3(x+2) +c = 3x c = -6

  5. Callisto
    • 2 years ago
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    Ha! Maybe I'm not even going in the right direction!

  6. thomas5267
    • 2 years ago
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    OK, but then the problem tells us to integrate it. \[ \begin{align*} &\int^3_2\frac{3x}{(x+2)^2+1}\,dx\\ =&\int^5_3\frac{3t-6}{t^2+1}\,dt \end{align*} \] As far as I can see, this cannot be integrated, or is my brain not functioning properly now.

  7. Callisto
    • 2 years ago
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    When x=2, t= 2+2 = 4... \[\int_4^5\frac{3t-6}{t^2+1}dt\]\[=\int_4^5\frac{3t}{t^2+1}-\frac{6}{t^2+1}dt\] For \(\int\frac{3t}{t^2+1}dt\) \[\int\frac{3t}{t^2+1}dt = \frac{3}{2}\int\frac{1}{t^2+1}d(t^2) \]Shouldn't be difficult For\(\int\frac{6}{t^2+1}dt\), use trigo sub.

  8. thomas5267
    • 2 years ago
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    OK, my brain is certainly not working properly. What is happening in here? I have never see something like this. \[ \int\frac{3t}{t^2+1}\,dt=\frac{3}{2}\int\frac{1}{t^2+1}\,d(t^2) \]

  9. Callisto
    • 2 years ago
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    d/dt (t^2) = 2t, agree?

  10. thomas5267
    • 2 years ago
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    I still don't understand \(d(t^2)\) and what should I do with it. I would do this in the following way. \[ \begin{align*} &\int\frac{3t}{t^2+1}\,dt\\ =&3\int\frac{t}{t^2+1}\,dt\\ =&\frac{3}{2}\int\frac{1}{u}\,du\\ =&\frac{3}{2}\ln(|t^2+1|)+C \end{align*} \]

  11. Callisto
    • 2 years ago
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    Basically, it's just substitution.. u = t^2 +1 du = 2t dt t dt = du/2 But you keep it as t, anyway, forget it! And you get it right.

  12. Callisto
    • 2 years ago
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    But remember it's definite integral.. not indefinite..

  13. thomas5267
    • 2 years ago
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    Can you teach me how to do this in your way? It seems way faster then mine.

  14. Callisto
    • 2 years ago
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    Hmm... I just did integration for the numerator and put it after ''d'' \[\int\frac{3t}{t^2+1}dt\]If you integrate the numerator, you'll get \(\frac{t^2}{2}\)So, \[\int\frac{3t}{t^2+1}dt = 3\int\frac{t}{t^2+1}dt =3 \int\frac{1}{t^2+1}d(\frac{t^2}{2})=\frac{3}{2} \int\frac{1}{t^2+1}d(t^2) \] Another way to think about it is that, d/dt (t^2 /2) gives you t, the numerator

  15. Callisto
    • 2 years ago
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    Did... I make it worse??

  16. thomas5267
    • 2 years ago
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    So I can just integrate the numerator? What should I do with \(\int\frac{1}{t^2+1}d(t^2)\) then?

  17. Callisto
    • 2 years ago
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    Get the answer! ln|t^2 +1| +C! Just consider t^2 = u, it's simply \(\int \frac{1}{u+1}du\) , that is ln |u+1|+C = ln|t^2+1|+C

  18. thomas5267
    • 2 years ago
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    Is the following true? \[ \int^5_4\frac{3t}{t^2+1}dt = \frac{3}{2}\int^{25}_{16}\frac{1}{t^2+1}d(t^2) \]

  19. Callisto
    • 2 years ago
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    No, you didn't change the variable, so you don't have to change the bound value.

  20. thomas5267
    • 2 years ago
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    What is the name of the procedure? I would like to find more information on the internet.

  21. Callisto
    • 2 years ago
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    I'm sorry.. I don't know the name of it :(

  22. thomas5267
    • 2 years ago
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    Anyways, thank you! I have learnt a lot in here.

  23. Callisto
    • 2 years ago
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    Welcome :)

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