# Thread: Integral of r^2 times square root of 4r^2 + 1 dr.

1. ## Integral of r^2 times square root of 4r^2 + 1 dr.

Can anyone help me with this?

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2. ## Re: Integral of r^2 times square root of 4r^2 + 1 dr.

Originally Posted by Frankkair
Can anyone help me with this?

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Let $\displaystyle u = 4r^2 + 1\;,\,du = 8rdr.\;\& \,\begin{array}{*{20}c} r &\vline & 0 & 2 \\\hline u &\vline & 1 & {17} \\ \end{array}$

Thus $\displaystyle \int_1^{17} {\left( {\frac{{u - 1}}{4}} \right)\sqrt u \left( {\frac{{du}}{8}} \right)}$.

3. ## Re: Integral of r^2 times square root of 4r^2 + 1 dr.

Thank you, but again I have a question (now a more basic one)

I got this, but I can't solve (guess my algebra is not working today)

Could u help me? Thanks!

4. ## Re: Integral of r^2 times square root of 4r^2 + 1 dr.

Seriously? Just put "17" and "1" in for u and subtract. You really can't do much with $\displaystyle 17^{5/2}$ or $\displaystyle 17^{3/2}$ but, of course, $\displaystyle 1^{5/2}= 1^{3/2}= 1$.