Question
Evaluate the following integrals:
$\int\sqrt{2\text{x}^2+3\text{x}+4}\text{dx}$
$\int\sqrt{2\text{x}^2+3\text{x}+4}\text{dx}$
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Answer
$\text{I}=\int\sqrt{2\text{x}^2+3\text{x}+4}\text{dx}$
$=\sqrt2\int\sqrt{\text{x}^2+\frac{3}{2}\text{x}+2}\text{dx}$
$=\sqrt2\int\sqrt{\text{x}^2+\frac{3}{2}\text{x}+\frac{9}{16}+\frac{23}{16}}\text{dx}$
$=\sqrt2\int\sqrt{\Big(\text{x}+\frac{3}{4}\Big)^2+\Big(\frac{\sqrt{23}}{4}\Big)^2}\text{dx}$
$=\sqrt2\begin{Bmatrix}\frac{\big(\text{x}+\frac{3}{4}\big)}{2}\sqrt{\text{x}^2+\frac{3}{2}\text{x}+2}+\frac{23}{32}\\\times\log\bigg|\Big(\text{x}+\frac{3}{4}\Big)+\sqrt{\text{x}^2+\frac{3}{2}\text{x}+2}\bigg|+\text{C}\end{Bmatrix}$
$\therefore\ \text{I}=\frac{4\text{x}+3}{8}\sqrt{2\text{x}^2+3\text{x}+4}+\frac{23\sqrt2}{32}\\\times\log\bigg|\Big(\text{x}+\frac{3}{4}\Big)+\sqrt{\text{x}^2+\frac{3}{2}\text{x}+2}\bigg|+\text{C}$
$=\sqrt2\int\sqrt{\text{x}^2+\frac{3}{2}\text{x}+2}\text{dx}$
$=\sqrt2\int\sqrt{\text{x}^2+\frac{3}{2}\text{x}+\frac{9}{16}+\frac{23}{16}}\text{dx}$
$=\sqrt2\int\sqrt{\Big(\text{x}+\frac{3}{4}\Big)^2+\Big(\frac{\sqrt{23}}{4}\Big)^2}\text{dx}$
$=\sqrt2\begin{Bmatrix}\frac{\big(\text{x}+\frac{3}{4}\big)}{2}\sqrt{\text{x}^2+\frac{3}{2}\text{x}+2}+\frac{23}{32}\\\times\log\bigg|\Big(\text{x}+\frac{3}{4}\Big)+\sqrt{\text{x}^2+\frac{3}{2}\text{x}+2}\bigg|+\text{C}\end{Bmatrix}$
$\therefore\ \text{I}=\frac{4\text{x}+3}{8}\sqrt{2\text{x}^2+3\text{x}+4}+\frac{23\sqrt2}{32}\\\times\log\bigg|\Big(\text{x}+\frac{3}{4}\Big)+\sqrt{\text{x}^2+\frac{3}{2}\text{x}+2}\bigg|+\text{C}$
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