Computing $\int \frac{6x}{\sqrt{x^2+4x+8}} dx$ I am trying to compute an indefinite integral. Thanks!
$$\int \frac{6x}{\sqrt{x^2+4x+8}} dx.$$
 A: First, try the substitution $u=x+2$, then use an appropriate trigonometric substitution.
A: $$\begin{align}\int \dfrac{6x}{\sqrt{x^2 + 4x + 8}}\, dx\;
& = \;\int \dfrac{6x+12 -12}{\sqrt{x^2 + 4x + 8}} \,dx \\ \\ \\ \\
& = \int \dfrac {6(x + 2) - 12}{\sqrt {\color{blue}{\bf (x^2+ 4x + 4) + 4}}}\,dx 
\end{align}$$
split the integral into the difference of two integrals:
$$ = \;\int \dfrac {6(x + 2)}{\sqrt {(x^2 + 4x + 8}}\,dx\;\; -\;\; 12\int \dfrac{1}{\sqrt{\color{blue}{\bf (x+2)^2 + 2^2}}}\, dx $$
Now, for the left hand integral, 


*

*let $u = x^2 + 4x + 8 \implies du = 2x + 4 = 2(x + 2) \,dx,\implies\; 3du = 6(x+2)\,dx$. Substitute: you'll have an integral of the form $\;3\int u^{-1/2} \,du$


For the right-hand integral, 


*

*let $\;v = x+2\;\implies\;dv = dx.$

*And as suggested, you'll want to use an appropriate trigonometric
substitution for the right-hand integral: in particular, find the
promising hyperbolic trig substitution.
A: If you have a quadratic expression in your integral, especially in the denominator or under a square root (or, in this case, both!), you should do the following steps:
1: Complete the square. Remember that this will look like $(x + a)^2 + b$, where $a$ is equal to one half the $x$ term, and $b$ is whatever you need to add to get the constant term to match up. In your case, the $x$ term is $4x$, so $a$ should be $1/2 \cdot 4 = 2$. So you get $(x + 2)^2 + b = x^2 + 4x + 8$, and so it's easy to see that $b$ should be $4$:
$$x^2 + 4x + 8 = (x + 2)^2 + 4$$
2: Make the substitution $u = x + 2$, to clean up the quadratic expression a bit. This isn't necessary, but I find it makes it nicer, and match up with other situations more easily. Your integral becomes:
$$\int \frac{6(u -2)}{\sqrt{u^2 + 4}}du$$
3: Now you should recognize this as a problem where you should use a trig substitution. The proper substitution to make in this case is:
$$u = 2\tan(\theta)$$
$$du = 2\sec^2(\theta)$$
The reason this is so nice is because:
$$\sqrt{u^2 + 4} = \sqrt{4\tan^2(\theta) + 4} = 2\sqrt{\tan^2(\theta) + 1} = 2\sqrt{\sec^2{\theta}} = \sec(\theta)$$
And so your integral becomes:
$$\int \frac{6(\tan(\theta) - 2)(2\sec^2(\theta))}{2\sec(\theta)}d\theta$$
$$= 6\int [\tan(\theta)\sec(\theta) - 2\sec(\theta)] d\theta$$
4: You should now be able to compute this integral using standard methods for integrating trig functions. For instance, the integral of $\tan(\theta)\sec(\theta)$ is $\sec(\theta)$- you should memorize this. The integral of $\sec(\theta)$ is something more complicated, but the formula is probably given (another good thing to memorize).
5: And so, you have your formula:
$$6\sec(\theta) - 12\ln(|\sec(\theta) + \tan(\theta)|) + C$$
Now you need to substitute $\theta = \tan^{-1}(u/2)$. You want to come up with a simple expression for $\sec(\tan^{-1}(u/2)$, the best way to do this is to draw a right triangle with the angle $\theta$ having a tangent of $u/2$. So the opposite leg has length $u$ and the adjacent leg has length $2$. Now, you can see that the secant of $\theta$ is $\sqrt{u^2 + 4}/2$. And so:
$$3\sqrt{u^2 + 4} - 12\ln\left|\frac{\sqrt{u^2 + 4}}{2} + \frac{u}{2}\right| + C$$
Finally, plug in $u = x + 2$:
$$3\sqrt{(x + 2)^2 + 4} - 12\ln\left|\frac{\sqrt{(x + 2)^2 + 4}}{2} + \frac{x + 2}{2}\right| + C$$
A: $$\int \frac{6x}{\sqrt{x^2+4x+8}}dx=\int\frac{6(x+2)-12}{\sqrt{(x+2)^2+2}}dx$$
$$=6\int\frac{x}{\sqrt{x^2+2}}dx-12\int\frac{1}{\sqrt{x^2+2}}dx$$
$$=6\sqrt{x^2+2}-12\sinh^{-1}\left(\frac{x+2}{2}\right)+C.$$
A: $$
\begin{align}
\int \frac{6x +12 -12}{\sqrt{x^2+4x+8}} dx
  &= \int \frac{6x +12}{\sqrt{x^2+4x+8}} dx - \int \frac{12}{\sqrt{x^2+4x+8}} dx \\
  &= \int \frac{6(x+2)}{\sqrt{x^2+4x+8}} dx - \int \frac{12}{\sqrt{x^2+4x+8}} dx \\ 
  &= \int \frac{3(2x+4)}{\sqrt{x^2+4x+8}} dx - \int \frac{12}{\sqrt{x^2+4x+8}} dx.
\end{align}$$
Then take 
$x^2+4x+8 = t$ substitution in 1st integral and performed perfect polynomial in second integral then simplify to get the answer.
A: Here is the Wolfram Alpha URL with the solution. http://www.wolframalpha.com/input/?i=integrate+6x+%2F+%28sqrt%28x%5E2+%2B+4x+%2B+8%29%29
