Euler substitution integrate $\int\sqrt{4x^2+5x+6} dx$

Question

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

Answer 1

HINT:

$$\int\sqrt{4x^2+5x+6}\space\text{d}x=$$ $$\int\sqrt{\left(2x+\frac{5}{4}\right)^2+\frac{71}{16}}\space\text{d}x=$$

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Substitute $u=2x+\frac{5}{4}$ and $\text{d}u=2\space\text{d}x$:

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$$\frac{1}{2}\int\sqrt{u^2+\frac{71}{16}}\space\text{d}x=$$

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Substitute $s=\arctan\left(\frac{4u}{\sqrt{71}}\right)$ and $\text{d}u=\frac{\sec^2(u)\sqrt{71}}{4}\space\text{d}s$:

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$$\frac{71}{32}\int\sec^3(s)\space\text{d}s=$$ $$\frac{61\tan\left(s\right)\sec\left(s\right)}{64}+\frac{71}{32}\int\sec(s)\space\text{d}s=$$ $$\frac{61\tan\left(s\right)\sec\left(s\right)}{64}+\frac{71\ln\left(\tan\left(s\right)+\sec\left(s\right)\right)}{64}+\text{C}=$$ $$\frac{61\tan\left(\arctan\left(\frac{4u}{\sqrt{71}}\right)\right)\sec\left(s\right)}{64}+\frac{71\ln\left(\tan\left(\arctan\left(\frac{4u}{\sqrt{71}}\right)\right)+\sec\left(\arctan\left(\frac{4u}{\sqrt{71}}\right)\right)\right)}{64}+\text{C}=$$ $$\frac{4(8x+5)\sqrt{4x^2+5x+6}+71\sinh^{-1}\left(\frac{8x+5}{\sqrt{71}}\right)}{64}+\text{C}$$

Answer 2

Note that $\sqrt{4x^2+5x+6}= \frac{\sqrt{71}}4 \sqrt{1+(\frac{8x+5}{\sqrt{71}})^2}$, permitting the substitution $\frac{8x+5}{\sqrt{71}}=\sinh t$ to integrate \begin{align} \int \sqrt{4x^2+5x+6}\ dx =\frac{{71}}{32} \int \cosh^2t\ dt = \frac{71}{128}\left(\sinh2t+2t \right)+C \end{align}

Answer 3

Hint : $4x^2+5x+6=4(x^2+5/4x+3/2)=4[x^2+2.x.5/8+25/64+71/64]=4[(x+5/8)^2+71/64]$

So , $\displaystyle \sqrt{4x^2+5x+6}=2.\sqrt{(x+\frac{5}{8})^2+71/64}$