Question

Integrate ( (x+5) / (5-x)) ^1/2 dx How???

Answer 1

http://openstudy.com/study#/updates/542f25a1e4b0fabfe0cab23b

Answer 2

The problem was explained there!

Answer 3

@Isaiah.Feynman, not the same problem
\[\int\sqrt{\frac{x+5}{5-x}}~dx\]
Let \(u=5-x\), then \(-du=dx\).
\[-\int\sqrt{\frac{10-u}{u}}~du\\
-\int\sqrt{\frac{10}{u}-1}~du\]
Let \(t=\dfrac{10}{u}\), then \(-dt=\dfrac{10}{u^2}~du\), or \(-\dfrac{10}{t^2}dt=du\).
\[10\int\frac{\sqrt{t-1}}{t^2}~dt\]
Integrate by parts, setting
\[\begin{matrix}
a=\sqrt{t-1}&&&db=\frac{dt}{t^2}\\
da=\frac{dt}{2\sqrt{t-1}}&&&b=-\frac{1}{t}
\end{matrix}\]
This gives you
\[10\left(-\frac{\sqrt{t-1}}{t}+\frac{1}{2}\int\frac{dt}{t\sqrt{t-1}}\right)\]
For the integral, let \(s=\sqrt{t-1}\), which gives \(s^2=t-1\) and \(2s~ds=dt\).
\[10\left(-\frac{\sqrt{t-1}}{t}+\frac{1}{2}\int\frac{2s}{(s^2+1)s}~ds\right)\\
10\left(-\frac{\sqrt{t-1}}{t}+\int\frac{ds}{s^2+1}\right)\]