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

**Frankkair** · February 14th 2013, 07:21 AM

Can anyone help me with this?

Integral of r^2 times square root of 4r^2 + 1 dr.-integral.jpg

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**Plato** · February 14th 2013, 07:42 AM

Originally Posted by Frankkair

Can anyone help me with this?

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

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

**Frankkair** · February 14th 2013, 08:01 AM

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)

Integral of r^2 times square root of 4r^2 + 1 dr.-problem.jpg

Could u help me? Thanks!

**HallsofIvy** · February 14th 2013, 10:51 AM

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

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