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anonymous

  • 5 years ago

what is the process of solving the indefinite integral x/sqrt(3x-1)dx

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  1. anonymous
    • 5 years ago
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    Integration by parts.

  2. anonymous
    • 5 years ago
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    or you could use u substitution

  3. anonymous
    • 5 years ago
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    I'm not sure if u-sub could work. By the way, salam. :P

  4. anonymous
    • 5 years ago
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    U sub doesnt work

  5. anonymous
    • 5 years ago
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    Use integration by parts. Bring sqrt(3x-1) on the top. So you will have something that looks like this.: \[x(3x-1)^{-1/2}\] Now, use integration by parts. u=x, dv=(3x-1)^(-1/2) Then use \[u.v-\int\limits_{}^{}v.du\]

  6. anonymous
    • 5 years ago
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    alaikum salam ... and I'm about to show how:\[\int\limits\limits_{} \frac{x}{\sqrt{3x-1}}dx \rightarrow u=3x-1, \frac{du}{3}=dx, x=\frac{u+1}{3}\] \[\frac{1}{3}\int\limits_{}\frac{x}{\sqrt{u}}du\] you have to back substitute x in terms of u \[\frac{1}{3}\int\limits_{}\frac{\frac{u+1}{3}}{\sqrt{u}}du \rightarrow \frac{1}{9}\int\limits_{}\frac{u+1}{\sqrt{u}}du\] \[\rightarrow \frac{1}{9}\int\limits_{}\frac{u}{\sqrt{u}}du+\frac{1}{9}\int\limits_{}\frac{1}{\sqrt{u}}du\] \[\rightarrow \frac {1}{9}\int\limits_{}\sqrt{u}du+\frac{1}{9}\int\limits_{}\frac{1}{\sqrt{u}}du\] \[\rightarrow \frac {2}{27}u^{\frac{3}{2}}+\frac{2}{9}u^{\frac{1}{2}}+C\] now substitute u back in, u=3x-1 \[\frac {2}{27}(3x-1)^{\frac{3}{2}}+\frac{2}{9}(3x-1)^{\frac{1}{2}}+C\]

  7. anonymous
    • 5 years ago
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    Oh, yeah! You're right. I totally forgot about that method.

  8. anonymous
    • 5 years ago
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    Thank you so much ! i have a final tomorow this really helped

  9. anonymous
    • 5 years ago
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    Integration by part does work but you will have to use u substitution on the second integral

  10. anonymous
    • 5 years ago
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    no problem

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