Question

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

Answer 1

Complete the square:


`\displaystyle\int√(x^2+6x)\,\mathrm dx=\int√((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√((3\sec y)^2-9)(3\sec y\tan y)\,\mathrm dy`

`\displaystyle9\int√(\sec^2y-1)\sec y\tan y\,\mathrm dy`

`\displaystyle9\int√(\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(\sec y+\tan y)/(\sec y+\tan y)\,\mathrm dy=\int(\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{√(x^2+6x)}3-\frac92\ln\left|\frac{x+3}3+\frac{√(x^2+6x)}3\right|+C`

`\displaystyle\frac{(x+3)√(x^2+6x)}2-\frac92\ln\left|\frac{x+3+√(x^2+6x)}3\right|+C`

`\displaystyle\frac{(x+3)√(x^2+6x)}2-\frac92\ln\left|x+3+√(x^2+6x)\right|+C`