Solving an integral I am having trouble with this integral:
$$\int_0^4\frac{x}{\sqrt{x^2+2x+25}}dx.$$
One obvious thing would be to complete the square: $x^2+2x+25=(x+1)^2+24$. But then, I don't know which substitution to use. Can anyone help? Thank you.
 A: I believe the substitution you're looking for is 
$x+1=2\sqrt6\tan\theta,x=2\sqrt6\tan\theta-1$.
This will reduce the denominator to
$\sqrt{(2\sqrt6\tan\theta)^2+24}=\sqrt{24\tan^2\theta+24}=\sqrt{24\sec^2\theta}=2\sqrt6\sec\theta$
I assume you'll be able to take it from there?
A: You will have, by substitution of $y=\frac{x+1}{\sqrt{24}}$,
$$\int_0^4\frac{x}{\sqrt{x^2+2x+25}}dx=\int_\frac{1}{\sqrt{24}}^1\frac{(\sqrt{24}y-1)}{\sqrt{y^2+1}}dy$$
that becomes
$$\int_\frac{1}{\sqrt{24}}^1\frac{(\sqrt{24}y-1)}{\sqrt{y^2+1}}dy=\sqrt{24}\int_\frac{1}{\sqrt{24}}^1\frac{y}{\sqrt{y^2+1}}dy-\int_\frac{1}{\sqrt{24}}^1\frac{1}{\sqrt{y^2+1}}dy.$$
This gives finally
$$\int_0^4\frac{x}{\sqrt{x^2+2x+25}}dx=\sqrt{48} - \sqrt{26} + {\rm arcsinh}\left(\frac{1}{\sqrt{24}}\right) -  {\rm arcsinh}(1).$$
A: We can do this in still another way (what I might refer to as the 'naive' way - set everything under the radical to a variable and go). You started with 
$$\int_0^4\frac{xdx}{\sqrt{x^2+2x+25}} = \int_0^4 \frac{xdx}{\sqrt{(x+1)^2 + 24}}$$
Let $u = (x+1)^2 + 24$, so that $du = (2x+2)dx$
Then $\displaystyle \int_0^4 \frac{xdx}{\sqrt{(x+1)^2 + 24}} = \frac{1}{2}\int_0^4 \frac{2x + 2}{\sqrt{(x+1)^2 + 24}}dx - \int_0^4 \frac{dx}{\sqrt{(x+1)^2+ 24}}$
The second integral is a standard arcsinh integral, and with the change of variables above the first integral becomes $\displaystyle \frac{1}{2}\int \frac{du}{\sqrt u}$
I always like it when an integral can be computed in many ways.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With
  Euler First Substitution
  $\ds{x = {25 - t^{2} \over 2\pars{t - 1}}}$:

\begin{align}
\int_{0}^{4}{x\,\dd x \over \root{x^{2} + 2x + 26}} & =
\int_{3}^{5}\bracks{{12 \over \pars{t - 1}^{2}} - {1 \over t - 1} -
{1 \over 2}}\,\dd t = 3 - \ln\pars{2} - 1 =
\bbx{2 - \ln\pars{2}}
\end{align}
