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Integral: (x+2)/sqrt (3x-1)

  • Thread starter polepole
  • Start date
[tex]\int[/tex][tex](X + 2)/\sqrt{3X-1}[/tex]

one of my attempts: u=3X-1 , du/3=dX , x = (u+1)/3

1/3[tex]\int[/tex][tex](u+7).du/\sqrt{u}[/tex] = 1/3[tex]\int[/tex][tex]u .du/\sqrt{u}[/tex] + 7/3 [tex]\int[/tex][tex]du/\sqrt{u}[/tex] = 1/3[tex]\int[/tex][tex]u .du/\sqrt{u}[/tex] + 14/3. sqrt{u} = stuck...

the answer should be 2/27 . (3X + 20) .[tex]\sqrt{3X-1}[/tex]
I'm not sure I'm a hundred percent following your work, but it seems that you are just making algebra errors.

The substitution you are intending to make is fine.

Then you write:

[tex] \frac{1}{3} \int \frac {(u+7)}{\sqrt{u}} du [/tex]

as the integral after substitution. But this isn't right; plug u back in here, you'll find that the coefficient at front should be something different. Then you seem to have taken that 7 out as though it were a constant, but you can't do that either. Anyway, you are on the right track with substituting u = 3x - 1. Just carefully redo the algebra.
I forgot to multiply with 1/3 from du/3. becomes:

1/9[tex]\int[/tex][tex](u+7).du/\sqrt{u}[/tex] = 1/9[tex]\int[/tex][tex]u .du/\sqrt{u}[/tex] + 7/9 [tex]\int[/tex][tex]du/\sqrt{u}[/tex] = 1/9[tex]\int[/tex][tex]u .du/\sqrt{u}[/tex] + [tex]14/3. \int \sqrt{u}[/tex]

but i've found the solution from there.
btw: if you split the integral you can use 7 as an a constant and put it in front of the intgr sign
Last edited:
Oh, I see what you did. Yes, the 7 was fine, just the multiplier was off.
Integrals of the type:

\int{R(x, \sqrt[n]{a x + b}) \, dx}

where [itex]R(u, v)[/itex] is a rational function of both of its arguments can always be reduced to integrals of rational functions with the substitution:

t = \sqrt[n]{a x + b}
So you would've used t=sqrt(3X-1) ?

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