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प्रश्न
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उत्तर
\[\int\sqrt{3 x^2 + 4x + 1} \text{ dx }\]
\[ = \sqrt{3}\int\sqrt{x^2 + \frac{4}{3}x + \frac{1}{3}}\text{ dx }\]
\[ = \sqrt{3}\int\sqrt{x^2 + \frac{4}{3}x + \left( \frac{2}{3} \right)^2 - \left( \frac{2}{3} \right)^2 + \frac{1}{3}} \text{ dx }\]
\[ = \sqrt{3}\int\sqrt{\left( x + \frac{2}{3} \right)^2 - \frac{4}{9} + \frac{1}{3}} \text{ dx }\]
\[ = \sqrt{3}\int\sqrt{\left( x + \frac{2}{3} \right)^2 - \left( \frac{1}{3} \right)^2} \text{ dx }\]
\[ = \sqrt{3} \left[ \frac{1}{2}\left( x + \frac{2}{3} \right)\sqrt{\left( x + \frac{2}{3} \right)^2 - \left( \frac{1}{3} \right)^2} - \frac{1}{2} \times \left( \frac{1}{3} \right)^2 \text{ ln } \left| \left( x + \frac{2}{3} \right) + \sqrt{\left( x + \frac{2}{3} \right)^2 - \left( \frac{1}{3} \right)^2} \right| + C \right] ....................\left[ \because \int \sqrt{x^2 - a^2} dx = \frac{1}{2}x\sqrt{x^2 - a^2} - \frac{1}{2} a^2 \text{ ln
}\left| x + \sqrt{x^2 - a^2} \right| + C \right]\]
\[ = \frac{1}{6}\left( 3x + 2 \right)\sqrt{3 x^2 + 4x + 1} - \frac{\sqrt{3}}{18}\text{ ln } \left| \left( x + \frac{2}{3} \right) + \sqrt{x^2 + \frac{4}{3}x + \frac{1}{3}} \right| + C\]
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` ∫ \sqrt{tan x} sec^4 x dx `
If \[\int\frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx\]
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\[\int\frac{x^3}{\sqrt{x^8 + 4}} \text{ dx }\]
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