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ChrisV Group Title

Find the indefinite integral. Please show work.

  • 2 years ago
  • 2 years ago

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  1. ChrisV Group Title
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    \[\int\limits_{?}^{?}(x+1)5^(x+1)^2\]

    • 2 years ago
  2. ChrisV Group Title
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    that (x+1)^2 is an exponent

    • 2 years ago
  3. nbouscal Group Title
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    \[ \int(x+1)5^{(x+1)^2}dx \]That?

    • 2 years ago
  4. ChrisV Group Title
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    yes

    • 2 years ago
  5. .Sam. Group Title
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    \[\Huge \int\limits_{}^{}(x+1)(5)^{(x+1)^2}\]

    • 2 years ago
  6. .Sam. Group Title
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    substitution u==x+1 du=dx \[\text{}=\int\limits 5^{u^2} u \, du\] substitution again t=u^2 dt=2udu \[\frac{1}{2}\int\limits 5^t \, dt\] \[\frac{5^t}{2 \ln (5)}+c\] \[\frac{5^{(x+1)^2}}{\log (25)}+c\]

    • 2 years ago
  7. .Sam. Group Title
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    ln=log

    • 2 years ago
  8. nbouscal Group Title
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    Okay, so u=x+1 and we have \[ \int u\cdot 5^{u^2}du=\int u\cdot e^{u^2\log5}du=\frac{1}{2\log 5}\int 2\log5u\cdot e^{u^2\log5}du=\frac{e^{u^2\log5}}{2\log 5}=\frac{5^{(x+1)^2}}{2\log5} \]

    • 2 years ago
  9. nbouscal Group Title
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    Oh, Sam's way is a little easier.

    • 2 years ago
  10. ChrisV Group Title
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    well the answer the book gives me looks like

    • 2 years ago
  11. ChrisV Group Title
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    \[1/2(5^{(x+1)^2}/\ln5)+C\]

    • 2 years ago
  12. nbouscal Group Title
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    Yep, that's the exact same thing as my answer and Sam's answer, just written slightly different.

    • 2 years ago
  13. ChrisV Group Title
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    i just dont understand how they get to that

    • 2 years ago
  14. nbouscal Group Title
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    Which part of the method Sam and I used did you not follow? (Sam's is a bit easier to follow because he does substitution twice and I only used it once)

    • 2 years ago
  15. ChrisV Group Title
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    i understand u substitution

    • 2 years ago
  16. ChrisV Group Title
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    5^u^2 = (ln5)u^2 right?

    • 2 years ago
  17. nbouscal Group Title
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    \[ 5^{u^2}=e^{(\log5)u^2} \]

    • 2 years ago
  18. nbouscal Group Title
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    \[ \int a^xdx=\int e^{x\log a}dx=\frac{1}{\log a}\int(\log a) e^{x\log a}dx=\frac{e^{x\log a}}{\log a}=\frac{a^x}{\log a} \\\int a^xdx=\frac{a^x}{\log a} \]

    • 2 years ago
  19. ChrisV Group Title
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    ok here is where im confused

    • 2 years ago
  20. ChrisV Group Title
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    how does 5^t become 5^t/ln5

    • 2 years ago
  21. ChrisV Group Title
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    in sams answer

    • 2 years ago
  22. nbouscal Group Title
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    He integrates it, using the formula I derived in my last comment.

    • 2 years ago
  23. ChrisV Group Title
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    well thats where im lost. I do not know how to integrate that

    • 2 years ago
  24. ChrisV Group Title
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    \[1/2\int\limits5^t dt\]

    • 2 years ago
  25. ChrisV Group Title
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    i would say that that is (ln5)t

    • 2 years ago
  26. ChrisV Group Title
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    and obviously im wrong lol

    • 2 years ago
  27. nbouscal Group Title
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    I just showed you how to integrate that in my comment, did you not see that? For \(a\) as any constant.

    • 2 years ago
  28. nbouscal Group Title
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    \[\int a^xdx=\int e^{x\log a}dx=\frac{1}{\log a}\int(\log a) e^{x\log a}dx=\frac{e^{x\log a}}{\log a}=\frac{a^x}{\log a} \\\int a^xdx=\frac{a^x}{\log a}\]

    • 2 years ago
  29. ChrisV Group Title
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    k ill write that down and try to wrap my brain around it

    • 2 years ago
  30. ChrisV Group Title
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    some things are just beyond me i guess

    • 2 years ago
  31. nbouscal Group Title
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    You just have to remember that \(e^{x^y}=e^{xy}\), which makes it so that \(a^x=(e^{\log a})^x=e^{x\log a}\)

    • 2 years ago
  32. nbouscal Group Title
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    Sorry, that first part should read \((e^x)^y=e^{xy}\)

    • 2 years ago
  33. ChrisV Group Title
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    luckily i have a 99.8 average in this class so if i miss this on the final it shouldnt hurt much >.<

    • 2 years ago
  34. nbouscal Group Title
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    Just revisit the rules for differentiating/integrating exponential and logarithmic functions. They come in handy for a lot of tricky integrals.

    • 2 years ago
  35. ChrisV Group Title
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    i try, i think my brain is overloaded. It is finals week

    • 2 years ago
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