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<p>I have a test Tues on a Review sheet of Integrals. In class we only got to the first 20 some of 40 integrals and so the professor e-mailed the answers to them but we needed to work out the proof. I can't seem to find the answer. I have a test on Tuesday on this review sheet and the Math Center is closed for Memorial Day Monday!</p>

<p>Problem: Integral of dx/(Sqrt[1+Sqrt[x]])</p>

<p>The Math Center tutor got 4/3 * (Sqrt[x] - 2) * Sqrt[1+Sqrt[x]]
I got 4/3 * ((1+Sqrt[x])^(3/2) - 2*Sqrt[1+Sqrt[x]]</p>

<p>My proof is:
Integral dx/(Sqrt[1+Sqrt[x]])<br>
u=1+Sqrt[x]
du=1/2Sqrt[x]dx
2<em>Sqrt[x]du=dx
2(u-1)du=dx
Integral dx/(Sqrt)dx
Integral u^(-1/2)</em>2(u-1)du
2 Integral u^(-1/2)<em>(u-1)du
2 Integral u^(1/2) - u^(-1/2) du
4/3</em>u^(3/2) - 2<em>u^(1/2)
4/3 * ((1+Sqrt[x])^(3/2) - 2</em>Sqrt[1+Sqrt[x]]</p>

<p>What am I doing wrong?</p>

2

<p>2 Integral u^(1/2) - u^(-1/2) du
4/3<em>u^(3/2) - 2</em>u^(1/2)</p>

<p>You didn't distribute the 2 to both u^(1/2) and u^(-1/2). Second line should read
4/3<em>u^(3/2) - 4</em>u^(1/2)</p>

<p>Then pull out a factor of 4/3*u^(1/2) [sometimes it's good to clean up while things are still in terms of u]</p>

<p>4/3<em>u^(1/2)</em>(u-3)</p>

<p>The substitute in u=Sqrt[x]+1 and get
4/3<em>Sqrt[Sqrt[x]+1]</em>(Sqrt[x]-2)</p>

<p>Which is what <a href="http://www.wolfram.integrator.com%5B/url%5D"&gt;www.wolfram.integrator.com&lt;/a> gives as an answer.</p>

3

<p>dilksy, you suck!!!</p>

<p>I just finished it and refreshed the page and saw you already posted a solution.</p>