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First Fundamental Theorem of Calculus

∫ 3 x 2+4 x+1...

Question

$\int \sqrt{3 x^{2} + 4 x + 1} d x$

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Solution

$\int \sqrt{3 x^{2} + 4 x + 1} d x = \sqrt{3} \int \sqrt{x^{2} + \frac{4}{3} x + \frac{1}{3}} d x = \sqrt{3} \int \sqrt{x^{2} + \frac{4}{3} x + {(\frac{2}{3})}^{2} - {(\frac{2}{3})}^{2} + \frac{1}{3}} d x = \sqrt{3} \int \sqrt{{(x + \frac{2}{3})}^{2} - \frac{4}{9} + \frac{1}{3}} d x = \sqrt{3} \int \sqrt{{(x + \frac{2}{3})}^{2} - {(\frac{1}{3})}^{2}} d x = \sqrt{3} [\frac{1}{2} (x + \frac{2}{3}) \sqrt{{(x + \frac{2}{3})}^{2} - {(\frac{1}{3})}^{2}} - \frac{1}{2} \times {(\frac{1}{3})}^{2} \ln |(x + \frac{2}{3}) + \sqrt{{(x + \frac{2}{3})}^{2} - {(\frac{1}{3})}^{2}}| + C] [∵ \int \sqrt{x^{2} - a^{2}} d x = \frac{1}{2} x \sqrt{x^{2} - a^{2}} - \frac{1}{2} a^{2} \ln |x + \sqrt{x^{2} - a^{2}}| + C] = \frac{1}{6} (3 x + 2) \sqrt{3 x^{2} + 4 x + 1} - \frac{\sqrt{3}}{18} \ln |(x + \frac{2}{3}) + \sqrt{x^{2} + \frac{4}{3} x + \frac{1}{3}}| + C$

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