## anonymous 5 years ago what is the process of solving the indefinite integral x/sqrt(3x-1)dx

1. anonymous

Integration by parts.

2. anonymous

or you could use u substitution

3. anonymous

I'm not sure if u-sub could work. By the way, salam. :P

4. anonymous

U sub doesnt work

5. 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 $u.v-\int\limits_{}^{}v.du$

6. 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 $\frac {2}{27}(3x-1)^{\frac{3}{2}}+\frac{2}{9}(3x-1)^{\frac{1}{2}}+C$

7. anonymous

Oh, yeah! You're right. I totally forgot about that method.

8. anonymous

Thank you so much ! i have a final tomorow this really helped

9. anonymous

Integration by part does work but you will have to use u substitution on the second integral

10. anonymous

no problem