anonymous
  • anonymous
what is the process of solving the indefinite integral x/sqrt(3x-1)dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Integration by parts.
anonymous
  • anonymous
or you could use u substitution
anonymous
  • anonymous
I'm not sure if u-sub could work. By the way, salam. :P

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anonymous
  • anonymous
U sub doesnt work
anonymous
  • anonymous
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 \[u.v-\int\limits_{}^{}v.du\]
anonymous
  • anonymous
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 \[\frac {2}{27}(3x-1)^{\frac{3}{2}}+\frac{2}{9}(3x-1)^{\frac{1}{2}}+C\]
anonymous
  • anonymous
Oh, yeah! You're right. I totally forgot about that method.
anonymous
  • anonymous
Thank you so much ! i have a final tomorow this really helped
anonymous
  • anonymous
Integration by part does work but you will have to use u substitution on the second integral
anonymous
  • anonymous
no problem

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