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First Fundamental Theorem of Calculus

∫ 2 x 2+3 x+4...

Question

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

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Solution

$\int \sqrt{2 x^{2} + 3 x + 4} d x = \sqrt{2} \int \sqrt{x^{2} + \frac{3}{2} x + 2} d x = \sqrt{2} \int \sqrt{x^{2} + \frac{3}{2} x + {(\frac{3}{4})}^{2} - {(\frac{3}{4})}^{2} + 2} d x = \sqrt{2} \int \sqrt{{(x + \frac{3}{4})}^{2} - \frac{9}{16} + 2} d x = \sqrt{2} \int \sqrt{{(x + \frac{3}{4})}^{2} + {(\frac{\sqrt{23}}{4})}^{2}} d x = \sqrt{2} [\frac{x + \frac{3}{4}}{2} \sqrt{{(x + \frac{3}{4})}^{2} + {(\frac{\sqrt{23}}{4})}^{2}} + \frac{23}{32} \ln |x + \frac{3}{4} + \sqrt{x^{2} + \frac{3}{2} x + 2}|] + C [∵ \int \sqrt{x^{2} + a^{2}} d x = \frac{1}{2} x \sqrt{x^{2} + a^{2}} - \frac{1}{2} a^{2} \ln |x + \sqrt{x^{2} + a^{2}}| + C] = \sqrt{2} [(\frac{4 x + 3}{8}) \sqrt{x^{2} + \frac{3}{2} x + 2} + \frac{23}{32} \ln |x + \frac{3}{4} + \sqrt{x^{2} + \frac{3}{2} x + 2}|] + C = (\frac{4 x + 3}{8}) \sqrt{2 x^{2} + 3 x + 4} + \frac{23 \sqrt{2}}{32} \ln |x + \frac{3}{4} + \sqrt{x^{2} + \frac{3}{2} x + 2}| + C$

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