Solving $\int \frac{6x^{3}+7x^2-12x+1}{\sqrt{x^2+4x+6}}dx$

I have previously asked here the following integral:$$\int \frac{6x^{3}+7x^2-12x+1}{\sqrt{x^2+4x+6}}dx.$$

In the meantime I have received an answer from Achille Hui here, but there are still some points that I don't understand from his answer.

At some point there is:

1. Set $$B(y)$$ to $$ay^2 + by + c$$, RHS($$*1$$) becomes $$(6-3a)y^3 + \cdots$$. We should fix $$a$$ to $$2$$.

2. Repeat this procedure once more, we find $$c$$ should be fixed to $$24$$.

I don't understand how does he fix these numbers? Does he let them equal $$0$$ to obtain the final result?

I have used a slightly different approach using a substitution. This makes the integral into a routine computational effort.

$$I = \int \frac{6x^{3}+7x^2-12x+1}{\sqrt{x^2+4x+6}}dx$$ $$= \int \frac{6x^{3}+7x^2-12x+1}{\sqrt{(x+2)^2+2}}dx$$ Let $$x + 2 = \sqrt2\tan\theta \implies dx = \sqrt2\sec^2\theta\ d\theta\text{ and } x=\sqrt2\tan\theta-2$$

$$\implies I = \int\frac{12\sqrt2\tan^3\theta -58\tan^2\theta+32\sqrt2\tan\theta+5}{\sqrt2\sec\theta}\sqrt2\sec^2\theta\ d\theta$$ $$= \int (12\sqrt2\tan^3\theta\sec\theta -58\tan^2\theta\sec\theta+32\sqrt2\tan\theta\sec\theta+5\sec\theta)\ d\theta$$

$$= \color{red}{12\sqrt2\int (\sec^2\theta-1)\tan\theta\sec\theta\ d\theta} - \color{green}{58\int(\sec^2\theta-1)\sec\theta\ d\theta} + 32\sqrt2\int\tan\theta\sec\theta\ d\theta + 5\int\sec\theta\ d\theta$$

$$= \color{red}{12\sqrt2\int (\sec^2\theta)\tan\theta\sec\theta\ d\theta -12\sqrt2\int\tan\theta\sec\theta\ d\theta} - \color{green}{58\int(\sec^2\theta)\sec\theta\ d\theta +\underline{58\int\sec\theta\ d\theta}} + 32\sqrt2\int\tan\theta\sec\theta\ d\theta + \underline{5\int\sec\theta\ d\theta}$$

$$= 12\sqrt2\left(\frac{\sec^3\theta}{3} -\sec\theta\right) - 58\int\sec^3\theta\ d\theta + 32\sqrt2\sec\theta + 63\int\sec\theta\ d\theta$$

$$= 4\sqrt2\sec^3\theta + 20\sqrt2\sec\theta - 58\int\sec^3\theta\ d\theta+ 63\int\sec\theta\ d\theta$$ $$= 4\sqrt2\sec^3\theta + 20\sqrt2\sec\theta \color{green}{- 29}\left(\sec\theta\tan\theta + \color{green}{\ln(\sec\theta+\tan\theta)}\right) + \color{green}{63\ln(\sec\theta+\tan\theta)}$$

Using $$\ln(z+\sqrt{z^2+1}) = \sinh^{-1}z$$, we get

$$I = 4\sqrt2\sec\theta\left(\sec^2\theta + 5 -\frac{29}{4\sqrt2}\tan\theta\right) + \color{green}{34\sinh^{-1}(\tan\theta)}$$

Using $$\tan\theta = \dfrac{x+2}{\sqrt2}\text{ and } \sec\theta = \dfrac{\sqrt{x^2+4x+6}}{\sqrt2}$$, we get

$$I = 4\sqrt{x^2+4x+6}\left(\frac{x^2+4x+6}{2} + 5 - \frac{29}{4\sqrt2}\frac{x+2}{\sqrt2}\right) + 34\sinh^{-1}\left(\frac{x+2}{\sqrt2}\right)$$ $$= \sqrt{x^2+4x+6}\left(2x^2+8x+12+20-\frac{29}{2}x-29\right) + 34\sinh^{-1}\left(\frac{x+2}{\sqrt2}\right)$$ $$= \boxed{\sqrt{x^2+4x+6}\left(2x^2-\frac{13}{2}x+3\right) + 34\sinh^{-1}\left(\frac{x+2}{\sqrt2}\right)}$$

• I am sorry, but I am getting stuck here: $$= \int (12\sqrt2\tan^3\theta\sec\theta -58\tan^2\theta\sec\theta+32\sqrt2\color{red}\tan\theta\sec\theta+5\sec\theta)\ d\theta$$ $$= 12\sqrt2\int (\sec^2\theta-1)\tan\theta\sec\theta\ d\theta - 58\int(\sec^2\theta-1)\sec\theta\ d\theta + 32\sqrt2\sec\theta + 5\int\sec\theta\ d\theta$$ Where does the $tanx$ go for this line? – James Warthington May 4 at 2:31
• @JamesWarthington the whole expression is under the integral sign and we have $\int \tan\theta \sec\theta\ d\theta = \sec\theta\$, so the next line has the $32\sqrt2\sec\theta$ – user1952500 May 4 at 3:12
• @JamesWarthington have added the additional step for clarity – user1952500 May 4 at 5:25
• Thank you so much for your detailed answer, I am stuck here, could you explain further? How does the 5 in the above line become 63 down below?$$= 12\sqrt2\int (\sec^2\theta-1)\tan\theta\sec\theta\ d\theta - 58\int(\sec^2\theta- 1)\sec\theta\ d\theta + 32\sqrt2\int\tan\theta\sec\theta\ d\theta + \color{red}{5}\int\sec\theta\ d\theta$$ $$= 12\sqrt2\left(\frac{\sec^3\theta}{3} -\sec\theta\right) - 58\int\sec^3\theta\ d\theta + 32\sqrt2\sec\theta + \color{red}{63}\int\sec\theta\ d\theta$$ – James Warthington May 5 at 1:50
• @JamesWarthington look at the underlined terms and colors in the edited answer now. The $58$ and $5$ coefficients get added. – user1952500 May 5 at 2:03