TITLE: Integrate $I=\int_{-1}^3\frac{\sqrt{x+5}}{(1+\sqrt{2x+3})^2}dx$
QUESTION [6 upvotes]: Compute
$$
I=\int_{-1}^{3}\frac{\sqrt{\vphantom{\large A}\,x + 5\,}\,}
{\left(1 + \sqrt{\vphantom{\large A}\,2x + 3\,}\,\right)^{2}}\,{\rm d}x
$$
Here's what I have tried:
Let
$t = 1 + \sqrt{\vphantom{\large A}\,2x + 3\,}\,$ then we have
${\rm d}x=\left(t - 1\right)\,{\rm d}t$. So
$$
I=\frac{1}{\sqrt{\vphantom{\large A}\,2\,}\,}
\int_{2}^{4}\frac{\left(t - 1\right)
\sqrt{\vphantom{\large A}\,\left(t - 1\right)^{2} + 7\,}}{t^{2}}\,{\rm d}t
$$
Now, I have stuck.....can anyone help me????? Thanks.

REPLY [3 votes]: The final result of your original integral is $$\frac{5}{\sqrt{2}}-3-\frac{\log (98)}{\sqrt{2}}+\sqrt{2} \log \left(4+\sqrt{2}\right)+\frac{9}{8} \log \left(9-4 \sqrt{2}\right)
\\=\frac{1}{2} \left(5 \sqrt{2}-6-\sqrt{2} \log \left(9-4 \sqrt{2}\right)+9 \tanh ^{-1}\left(1-\frac{1}{\sqrt{2}}\right)\right)$$
according to Mathematica (the two forms are numerically equal). You can attempt to calculate this as follows: in your second expression, substitute $x=\sqrt{(t-1)^2+7}-t$, $t = \frac{8-x^2}{2x+2}$. This will turn the integrand into a rational function of $t$, which you can always integrate (in principle) by partial fractions. I hope for you this isn't homework, because it will likely be ugly.