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anonymous

  • 5 years ago

Evaluate the integral ∫ [x^3/sqrt(x^2+1)] in two different ways:? a) Integration by parts b) The substitution u = sqrt(x^2+1)

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  1. anonymous
    • 5 years ago
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    b)\[1/2\int\limits_{}^{}(u^{2}-1/u) du\]

  2. anonymous
    • 5 years ago
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    i think i know part A ) \[\int\limits x ^{3}\div \sqrt{x ^{2}+1}\] you can say that sqrt of x^2 + 1 is (x^2 + 1)^1/2 the answer will be > [x^4/4] / (x^2+1)^1/1 btw am not sure ><"

  3. anonymous
    • 5 years ago
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    a) \[uv-\int\limits_{}^{}vdv\] it is a a very long problem, I wouldn't do it by parts and I could see your professor really want you to do it on HW, but it will unlikely on your test.

  4. anonymous
    • 5 years ago
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    hey dumbcow i just need to ask you one more question about the question i put up before. when your free lol:)

  5. dumbcow
    • 5 years ago
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    ok

  6. dumbcow
    • 5 years ago
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    i may be wrong, but i dont see how audia4 got the answer for part b) \[u = \sqrt{x^{2}+1}\] \[du =\frac{x}{\sqrt{x^{2}+1}} dx\] \[\ dx=\frac{\sqrt{x^{2}+1}}{x} du\] replacing dx cancels out sqrt(x^2+1) and one x on top leaving x^2 du \[x^{2} = u^{2} -1\] so integral becomes \[\int\limits_{}^{}u^{2} -1 du\]

  7. anonymous
    • 5 years ago
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    i was wrong on b), wasnt thinking, sorry guys! dumbcow got the right answer!!!!

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spraguer (Moderator)
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