can you help me with integrate this math problem $$\int\sqrt{4x^2+5x+6}\, dx,$$ I tried with $1$. Euler substitution, but I stopped on this.
$$\int\frac{t^2-6}{5-4t}\cdot\frac{-2t^2+5t-12}{5-4t}\cdot\frac{10t-4t^2-24}{(5-4t)^2} dt$$
Can you help what will be next step or any better solution?
Thank you
HINT:
$$\int\sqrt{4x^2+5x+6}\space\text{d}x=$$ $$\int\sqrt{\left(2x+\frac{5}{4}\right)^2+\frac{71}{16}}\space\text{d}x=$$
Substitute $u=2x+\frac{5}{4}$ and $\text{d}u=2\space\text{d}x$:
$$\frac{1}{2}\int\sqrt{u^2+\frac{71}{16}}\space\text{d}x=$$
Substitute $s=\arctan\left(\frac{4u}{\sqrt{71}}\right)$ and $\text{d}u=\frac{\sec^2(u)\sqrt{71}}{4}\space\text{d}s$:
$$\frac{71}{32}\int\sec^3(s)\space\text{d}s=$$ $$\frac{61\tan\left(s\right)\sec\left(s\right)}{64}+\frac{71}{32}\int\sec(s)\space\text{d}s=$$ $$\frac{61\tan\left(s\right)\sec\left(s\right)}{64}+\frac{71\ln\left(\tan\left(s\right)+\sec\left(s\right)\right)}{64}+\text{C}=$$ $$\frac{61\tan\left(\arctan\left(\frac{4u}{\sqrt{71}}\right)\right)\sec\left(s\right)}{64}+\frac{71\ln\left(\tan\left(\arctan\left(\frac{4u}{\sqrt{71}}\right)\right)+\sec\left(\arctan\left(\frac{4u}{\sqrt{71}}\right)\right)\right)}{64}+\text{C}=$$ $$\frac{4(8x+5)\sqrt{4x^2+5x+6}+71\sinh^{-1}\left(\frac{8x+5}{\sqrt{71}}\right)}{64}+\text{C}$$