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### Answers (1)

Answer:-

$\frac{4 x+3}{8} \sqrt{2 x^{2}+3 x+4}+\frac{23 \sqrt{2}}{32} \log \left|x+\frac{3}{4} \sqrt{x^{2}-\frac{3}{2}+2}\right|+c$

Hint:-

Multiplying by  $2\sqrt{2}$

Given:-

$\int \sqrt{2 x^{2}+3 x+4} d x$

Solution:-

\begin{aligned} &=\frac{1}{2 \sqrt{2}} \int \sqrt{16 x^{2}+24 x+32} d x \\\\ &=\frac{1}{2 \sqrt{2}} \int \sqrt{(4 x)^{2}+2.4 x \cdot 3+(3)^{2}+23} d x \\\\ &=\frac{1}{2 \sqrt{2}} \int \sqrt{(4 x+3)^{2}+\sqrt{23}^{2}} d x \end{aligned}

By using the formula

\begin{aligned} &\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \\\\ &=\frac{1}{2 \sqrt{2}}\left[\frac{1}{2} \times \frac{1}{4} \times(4 x+3) \sqrt{(4 x+3)^{2}}+\frac{\sqrt{23}^{2}}{2} \times \frac{1}{4} \log \left[(4 x+3)+\sqrt{(4 x+3)^{2}+\sqrt{23}}\right]+c\right] \end{aligned}

\begin{aligned} &=\frac{4 x+3}{8} \sqrt{2 x^{2}+3 x+4}+\frac{23 \sqrt{2}}{32} \log \left|(4 x+3)+\sqrt{2 x^{2}+3 x+4}\right|+c \\\\ &=\frac{4 x+3}{8} \sqrt{2 x^{2}+3 x+4}+\frac{23 \sqrt{2}}{32} \log \left|\left(x+\frac{3}{4}\right) \sqrt{x^{2}-\frac{3}{2}+2}\right|+\frac{23 \sqrt{2}}{32} \log 4+c \\\\ &=\frac{4 x+3}{8} \sqrt{2 x^{2}+3 x+4}+\frac{23 \sqrt{2}}{32} \log \left|x+\frac{3}{4} \sqrt{x^{2}-\frac{3}{2}+2}\right|+c \end{aligned}

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