Question

How do you evaluate sqrt(x^2 +6x)dx

Answer 1

Complete the square:

$\displaystyle\int\sqrt{x^2+6x}\,\mathrm dx=\int\sqrt{(x+3)^2-9}\,\mathrm dx$

Now substitute $x+3=3\sec y$, so that $\mathrm dx=3\sec y\tan y\,\mathrm dy$. The integral then becomes

$\displaystyle\int\sqrt{(3\sec y)^2-9}(3\sec y\tan y)\,\mathrm dy$
$\displaystyle9\int\sqrt{\sec^2y-1}\sec y\tan y\,\mathrm dy$
$\displaystyle9\int\sqrt{\tan^2y}\sec y\tan y\,\mathrm dy$
$\displaystyle9\int\sec y\tan^2y\,\mathrm dy$

Use the Pythagorean theorem to reduce the integrand to

$\displaystyle9\int\sec y(\sec^2y-1)\,\mathrm dy$
$\displaystyle9\int(\sec^3y-\sec y)\,\mathrm dy$

You can integrate $\sec^3y$ by parts, setting

$\begin{matrix}u=\sec y&&\mathrm dv=\sec^2y\,\mathrm dy\\\mathrm du=\sec y\tan y\,\mathrm dy&&v=\tan y\end{matrix}$

So,

$\displaystyle\int\sec^3y\,\mathrm dy=\sec y\tan y-\int\sec y\tan^2y\,\mathrm dy$
$\displaystyle2\int\sec^3y\,\mathrm dy=\sec y\tan y+\int\sec y\,\mathrm dy$
$\displaystyle\int\sec^3y\,\mathrm dy=\frac12\sec y\tan y+\frac12\int\sec y\,\mathrm dy$

This means you have

$\displaystyle9\left(\frac12\sec y\tan y+\frac12\int\sec y\,\mathrm dy\right)-9\int\sec y\,\mathrm dy$
$\displaystyle\frac92\sec y\tan y-\frac92\int\sec y\,\mathrm dy$

You can integrate $\sec y$ by writing

$\displaystyle\int\sec y\frac{\sec y+\tan y}{\sec y+\tan y}\,\mathrm dy=\int\frac{\sec^2y+\sec y\tan y}{\sec y+\tan y}\,\mathrm dy=\int\frac{\mathrm dz}z=\ln|\sec y+\tan y|+C$

So you are left with

$\displaystyle\frac92\sec y\tan y-\frac92\ln|\sec y+\tan y|+C$

Transforming back to $x$ gives you

$\displaystyle\frac92\sec\left(\arcsec\frac{x+3}3\right)\tan\left(\arcsec\frac{x+3}3\right)-\frac92\ln\left|\sec\left(\arcsec\frac{x+3}3\right)+\tan\left(\arcsec\frac{x+3}3\right)\right|+C$
$\displaystyle\frac92\frac{x+3}3\frac{\sqrt{x^2+6x}}3-\frac92\ln\left|\frac{x+3}3+\frac{\sqrt{x^2+6x}}3\right|+C$
$\displaystyle\frac{(x+3)\sqrt{x^2+6x}}2-\frac92\ln\left|\frac{x+3+\sqrt{x^2+6x}}3\right|+C$
$\displaystyle\frac{(x+3)\sqrt{x^2+6x}}2-\frac92\ln\left|x+3+\sqrt{x^2+6x}\right|+C$