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#### paulmdrdo

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- May 13, 2013

- 386

3.) ∫(3+s)

^{1/2}(s+1)^{2}ds- Thread starter paulmdrdo
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- May 13, 2013

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3.) ∫(3+s)^{1/2}(s+1)^{2}ds

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- May 13, 2013

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how do you get (u-2)I would let:

\(\displaystyle u=s+3\,\therefore\,du=ds\)

and now we have:

\(\displaystyle \int u^{\frac{1}{2}}(u-2)^2\,du\)

Now, expand, distribute, and then apply the power rule term by term.

- Feb 5, 2012

- 1,621

An alternative method without using substitutions is to write the integrand only using \(s+3\).

\begin{eqnarray}

\int(s+3)^{1/2}(s+1)^2\,ds&=&\int(s+3)^{1/2}(s+3-2)^2\,ds\\

&=&\int(s+3)^{1/2}\left((s+3)^2-4(s+3)+4\right)\,ds\\

&=&\int(s+3)^{5/2}\,d(s+3)-4\int(s+3)^{3/2}\,d(s+3)+4\int(s+3)^{1/2}\,d(s+3)\\

\end{eqnarray}

Hope you can continue.

- Feb 5, 2012

- 1,621

Hi paulmdrdo,how do you get (u-2)^{2}?

The \(s+1\) in the integrand becomes \(u-2\). That is, \(u=s+3\Rightarrow u-2=s+1\).