Can you convert (s+1) to (u-2) in the integration of (3+s)^(1/2)(s+1)^2ds?

[paulmdrdo1]

3.) ∫(3+s)^1/2(s+1)²ds

 

[MarkFL]

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.

 

[paulmdrdo1]

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.

how do you get (u-2)²?

 

[Sudharaka]

Hi everyone, :)

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. :)

 

[Sudharaka]

how do you get (u-2)²?

Hi paulmdrdo, :)

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