# Where have I made my mistake in my integrating $\int x \sqrt{2x - 1} dx$?

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

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.

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

$$\int x \sqrt{u} dx\tag{1}$$

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

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:

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

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

• @RossMillikan, it is certainly progress... now it is $\frac{1}{4} \int u^{3/2} du + \frac{1}{4} \int u^{1/2} du$. – vonbrand Mar 28 '13 at 22:51

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 $u$

• I was just noting that this substitution had not been given, and as I was starting to answer, your answer appeared. (+1) – robjohn Mar 28 '13 at 22:14

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 $$\int\frac{1}{4}(u+1)u^{1/2}du$$