Solving the integrand $\frac{x^3}{\sqrt{x^2+10x+16}}$ I just wanted to make sure that what I did to integrate $\frac{x^3}{\sqrt{x^2+10x+16}}$ is correct.
I assumed that it is classified as a trigonometric substitution problem. And so, what I first did is to apply "completing the square":
$\int \frac{x^3 dx}{\sqrt{x^2+10x+16}} = \int \frac{x^3 dx}{\sqrt{x^2+10x+25+16-25}} = \int \frac{x^3 dx}{\sqrt{(x+5)^2-9}}
$
After that, I assigned values into some variables:
let $a = 3 $
$ x + 5 = 3 \sec \Theta \rightarrow \sec \Theta = \frac{x+5}{3} $
$ dx = 3 \sec   \Theta \tan \Theta\, d \Theta$
Next, I have substituted the value of (x+5) to $\sqrt{(x+5)^2-9}$ which leads to $3 \tan \Theta$. And, $\tan \Theta$ is equal to $\frac{\sqrt{x^2+10x+16}}{3}$.
Afterwards, I have replaced all of the variables to the values that I assigned to them:
$\int \frac{x^3}{\sqrt{x^2+10x+16}} \rightarrow \int \frac{(3 \sec \Theta)^3 (3 \sec \Theta \tan \Theta)\, d\Theta}{3 \tan \Theta} \rightarrow \int (3 \sec\Theta - 5)^3 \sec\Theta\, d \Theta$
Expanding the trinomial, distributing $\sec \theta$ to each term and applying the constant theorem will result to:
$27\int \sec^4\Theta d\Theta - 135\int sec^3\Theta\, d\Theta + 225 \int sec^2\Theta d\Theta - 125 \int sec \Theta d \Theta$
Now, by making $\sec^2 \Theta$ into $(1 + \tan^2 \Theta)$ and applying u-substitution: $27 \int \sec^4 \Theta \,d\Theta \rightarrow 27 \tan \Theta + 9 \tan^3 \Theta + C$
By applying integration by parts, $\int \sec^3 \Theta\, d\Theta$ would become $\frac{\sec\Theta tan\Theta + \ln\left | \sec\Theta + \tan\Theta \right |}{2}$.
Finally, $\int \sec^2 \Theta\, d\Theta$ would simply become $\tan \Theta$ and $\int \sec \Theta$ would be $ln\left | \sec\Theta + \tan\Theta \right |$ .
Since $\sec \Theta$ is equal to $\frac{x+5}{3}$ and $\tan\Theta$ is equal to $\frac{\sqrt{x^2+10x+16}}{3}$, the whole integral would be (I had added all like terms before this):
$9\left [ \frac{\sqrt{x^2+10x+16}}{3} \right ]^3 + 252\left ( \frac{\sqrt{x^2+10x+16}}{3}\right ) -\frac{135}{2} \left ( \frac{x+5}{3}\right ) \left ( \frac{\sqrt{x^2+10x+16}}{3}\right ) - \frac{385}{2} ln \left | \frac{x+5}{3} + \frac{\sqrt{x^2+10x+16}}{3}\right | + C$
Lastly, I simplified the integral:
$\frac{1}{3} \left ( x^2 + 10x+16 \right )^\frac{3}{2} + \frac{(93-15x)\sqrt{x^2+10x+16}}{2} - \frac{385}{2} ln \left | \frac{x+5+\sqrt{x^2+10x+16}}{3}\right | + C$
Have I integrated the integrand appropriately? Thanks in advanced.
 A: It is correct. Here is another method without trigonometric substitution:
$$\int\frac{x^3}{\sqrt{x^2+10x+16}}dx=\int\frac{x^3+10x^2+16x-10x^2-100x-160+84x+160}{\sqrt{x^2+10x+16}}dx$$
$$ = \int\frac{(x-10)(x^2+10x+16)+84x+160}{\sqrt{x^2+10x+16}}dx$$
$$ = \int(x-10)\sqrt{x^2+10x+16}\ dx+\int\frac{84x+420}{\sqrt{x^2+10x+16}}dx-260\int\frac{1}{\sqrt{x^2+10x+16}}dx$$
The last integral can be directly solved using formula, second last one will follow if you substitute $t=x^2+10x+16$, remains to solve the first integral.
$$\int(x-10)\sqrt{x^2+10x+16}\ dx = \frac{1}{2}\int(2x+10)\sqrt{x^2+10x+16}\ dx-15\int\sqrt{x^2+10x+16}\ dx$$
Last integral follows directly from formula and second last is solved by substituting $x^2+10x+16$.
Note that this is a general method. In case there is any polynomial in numerator and any quadratic (with or without square root) in denominator, divide the numerator by denominator to get a linear in numerator, then reduce the numerator to a constant times the derivative of quadratic plus another constant, which can be solved easily using substitution and existing formulas.
