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$$\int x \sqrt{2x - 1} \,dx$$

Let $u = 2x - 1$

$$\int x \sqrt{u}\, dx$$

$$\frac{du}{dx} = 2 \implies \frac{1}{2}du = dx.$$

So the integral is written as

$$\int \frac{1}{2} u^{\frac{1}{2}} \, du$$ $$ = \frac{1}{2} \left(\frac{2}{3} u^{\frac{3}{2}} \right)$$ $$ = \frac{1}{3} (2x - 1)^{\frac{3}{2}} + c$$

But apparently this is wrong.

Wolfram says it is something completely different. Where have I made the mistake?

EDIT: OOOPS!! Sorry. Silly mistake.

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    $\begingroup$ Where did $x$ go from your second integral? $\endgroup$ – Thomas Andrews Mar 28 '13 at 21:51
  • $\begingroup$ @ThomasAndrews Whoops. I thought because I saw the $x$ in the $u$ bit, it for some reason carried without reading it properly. Oops. $\endgroup$ – Kaish Mar 28 '13 at 21:52
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$$\int x \sqrt{u} dx\tag{1}$$

$$\int \frac{1}{2} u^{\frac{1}{2}} du\tag{2?}$$

What happened to the factor of $x$ going from $(1) \to (2)$?

We have that $u = 2x-1$, so $x = \dfrac{u+1}{2},\;$ and as you know, $dx = \dfrac 12 du.\,$ This gives us:

$$\int x \sqrt{2x - 1} \,dx \quad = \quad \frac{1}{2} \int \frac{(u+1)}{2} u^{1/2} \, du\tag{2} \quad =\quad \frac 14 \int \left(u^{3/2} + u^{1/2}\right) \,du$$

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If $u = 2x-1$ then $x = \frac{u+1}{2}$ and so $$\int x \sqrt{2x-1}dx = \int \frac{1}{2}\frac{u+1}{2}\sqrt{u}du$$

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  • $\begingroup$ @RossMillikan, it is certainly progress... now it is $\frac{1}{4} \int u^{3/2} du + \frac{1}{4} \int u^{1/2} du$. $\endgroup$ – vonbrand Mar 28 '13 at 22:51
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I'd do it like this: $$ u = \sqrt{2x-1} $$ $$ u^2 = 2x-1 $$ $$ 2u\,du = 2\,dx $$ $$ u\,du = dx $$ $$ x = \frac{u^2+1}{2} $$ $$ \int x\sqrt{2x-1}\,dx = \int\frac{u^2+1}{2}\, u \,\Big(u\,du\Big) $$ Then you're integrating a polynomial function. When you've done that, change it back to a function of $x$ by putting $\sqrt{2x-1}$ in place of $u$

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  • $\begingroup$ I was just noting that this substitution had not been given, and as I was starting to answer, your answer appeared. (+1) $\endgroup$ – robjohn Mar 28 '13 at 22:14
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You dropped your $x$ when you went from $$\int x\sqrt{u}dx \implies \int\frac{1}{2}\sqrt{u}du $$ Since $x=\frac{1}{2}(u+1)$, the correct expression would be $$\int\frac{1}{4}(u+1)u^{1/2}du$$

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