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My problem, is how to find the integral of $$ \int \frac{2x}{\sqrt {4x-1}}\, \mathrm{d}x$$

Can I do this? $$ \int \frac{\frac{1}{2}\cdot4x-1+1}{\sqrt {4x-1}}\, \mathrm{d}x$$ $$ \frac{1}{2}\int \frac{4x-1+1}{\sqrt {4x-1}}\, \mathrm{d}x$$ $$ \frac{1}{2}\int \frac{4x-1}{\sqrt {4x-1}}+\frac{1}{\sqrt {4x-1}}\, \mathrm{d}x$$ $$ \frac{1}{2}\int \sqrt {4x-1}+\frac{1}{\sqrt {4x-1}}\, \mathrm{d}x$$ then $$ 4x-1= u/'$$ $$ 4dx= du$$ $$ dx= \frac{1}{4}du$$ insert that in the integral $$ \frac{1}{2}\int \sqrt {u}+\frac{1}{\sqrt {u}}\cdot\frac{1}{4}\, \mathrm{d}u$$ $$ \frac{1}{8}\int \sqrt {u}+\frac{1}{\sqrt {u}}\, \mathrm{d}u$$

and I get $$\frac{1}{12}\cdot u\cdot \sqrt{u}+ \frac{1}{4}\cdot \sqrt{u}+C$$ $$\frac{1}{12}\cdot(4x-1)\cdot \sqrt{4x-1}+ \frac{1}{4}\cdot \sqrt{4x-1}+C$$

and that's the correct solution. But can I just rewrite $2x$ as $\frac{1}{2}\cdot4x-1+1$ and then put $\frac{1}{2}$ before the integral? Thank you in advance!

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    $\begingroup$ Why not doing $\sqrt{4x-1}=2t$, so $4x-1=4t^2$, $x=t^2+\frac{1}{4}$ and $dx=2t\,dt$? Your method is good, but you should write $\frac{1}{2}(4x-1+1)$ $\endgroup$ – egreg Jan 7 '14 at 21:37
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    $\begingroup$ wolframalpha.com $\endgroup$ – chharvey Jan 7 '14 at 21:41
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You've got the right idea, but there are two places you erred:

$$ \int \frac{\frac{1}{2}\cdot4x-1+1}{\sqrt {4x-1}}\, \mathrm{d}x = \int \frac {2x - 1 + 1}{\sqrt{4x - 1}}\,dx \neq \frac{1}{2}\int \frac{4x-1+1}{\sqrt {4x-1}}\, \mathrm{d}x$$

You need to have: $$\int \dfrac{2x}{\sqrt{4x-1}}\,dx = \int \frac{\frac{1}2(4x-1) +\frac 12}{\sqrt {4x-1}}\, \mathrm{d}x = \frac 12\int \frac{(4x - 1) + 1}{\sqrt{4x-1}}$$

Here you also erred: $$ \frac{1}{2}\int\sqrt {u}+\frac{1}{\sqrt {u}}\cdot\frac{1}{4}\, \mathrm{d}u = \int \left(\frac 12 \sqrt u + \frac 1{8\sqrt u}\right)\,du \neq \frac{1}{8}\int \sqrt {u}+\frac{1}{\sqrt {u}}\, \mathrm{d}u $$

If you got the correct answer, it's because the second error reversed the first error.

If you go back and fix the first error, and work from there with the same idea (just be careful to note the scope of a a factor, and to distribute the factor, when needed!) you should obtain the correct result, and do so correctly.

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  • $\begingroup$ If I write it this way, will it be correct? $$ \frac{1}{2}\int (\sqrt {u}+\frac{1}{\sqrt {u}})\cdot\frac{1}{4}\, \mathrm{d}u$$ $$ \frac{1}{2}(\int \sqrt {u}\cdot\frac{1}{4}\, \mathrm{d}u+ \int \frac{1}{\sqrt {u}}\cdot\frac{1}{4}\, \mathrm{d}u)$$ and then $$ \frac{1}{2}(\frac{1}{4}\int \sqrt {u}\, \mathrm{d}u+ \frac{1}{4}\int \frac{1}{\sqrt {u}}\cdot\, \mathrm{d}u)$$ $$ \frac{1}{8}\int \sqrt {u}\, \mathrm{d}u+ \frac{1}{8}\int \frac{1}{\sqrt {u}}\cdot\, \mathrm{d}u$$ $\endgroup$ – L_McClain Jan 7 '14 at 22:13
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    $\begingroup$ I'd need to see your earlier work, but if that gets to you the top line, and is correct, then indeed the last line follows from the top line. $\endgroup$ – Namaste Jan 7 '14 at 22:16
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    $\begingroup$ Yes, it is correct. $\endgroup$ – Namaste Jan 7 '14 at 22:18
  • $\begingroup$ That's great! Thank you! I'm always happy to see your answers! :) $\endgroup$ – L_McClain Jan 7 '14 at 22:21
  • $\begingroup$ @amWhy: Looks like lots of green for u today. =1 $\endgroup$ – Amzoti Jan 8 '14 at 1:35
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You can't do in that way, but you can salvage the idea: $$ \int\frac{2x}{\sqrt{4x-1}}dx=\frac{1}{2}\int\frac{4x}{\sqrt{4x-1}}dx =\frac{1}{2}\int\frac{4x-1+1}{\sqrt{4x-1}}dx $$ and go on.

A possibly simpler strategy is doing the substitution $$ \sqrt{4x-1}=2t $$ where I use $2t$ because after squaring we get $$ 4x-1=4t^2 $$ so $$ x=t^2+\frac{1}{4},\quad dx=2t\,dt $$ and the integral becomes $$ \int\frac{2t^2+\frac{1}{2}}{t}2t\,dt=\int(4t^2+1)\,dt=\frac{4}{3}t^3+t+c $$ and the result will follow by substituting back $x$.

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  • $\begingroup$ So if I do in like you did in this answer, then it will be correct? $\endgroup$ – L_McClain Jan 7 '14 at 21:45
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    $\begingroup$ You have to be more careful when transporting factors. ;-) $\endgroup$ – egreg Jan 7 '14 at 22:05

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