int frac{2x^2-1+x^2sqrt{x^2+4}}{x^2(x^2+4)} d | vedclass

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(B) हम समाकल्य को इस प्रकार सरल करते हैं:$\int \frac{2x^2-1+x^2\sqrt{x^2+4}}{x^2(x^2+4)} dx = \int \left( \frac{2x^2}{x^2(x^2+4)} - \frac{1}{x^2(x^2+4

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$\frac{9}{8} 	an^{-1} \frac{x}{2} + \frac{1}{4x} + \sinh^{-1} \frac{x}{2} + c$

Vedclass · Answer

(B) हम समाकल्य को इस प्रकार सरल करते हैं:$\int \frac{2x^2-1+x^2\sqrt{x^2+4}}{x^2(x^2+4)} dx = \int \left( \frac{2x^2}{x^2(x^2+4)} - \frac{1}{x^2(x^2+4)} + \frac{x^2\sqrt{x^2+4}}{x^2(x^2+4)} ight) dx$$= \int \left( \frac{2}{x^2+4} - \frac{1}{x^2(x^2+4)} + \frac{1}{\sqrt{x^2+4}} ight) dx$$\frac{1}{x^2(x^2+4)} = \frac{1}{4} \left( \frac{1}{x^2} - \frac{1}{x^2+4} ight)$ के लिए आंशिक भिन्न का उपयोग करते हुए:$= \int \left( \frac{2}{x^2+4} - \frac{1}{4} \left( \frac{1}{x^2} - \frac{1}{x^2+4} ight) + \frac{1}{\sqrt{x^2+4}} ight) dx$$= \int \left( \frac{2}{x^2+4} + \frac{1}{4(x^2+4)} - \frac{1}{4x^2} + \frac{1}{\sqrt{x^2+4}} ight) dx$$= \int \left( \frac{9}{4(x^2+4)} - \frac{1}{4x^2} + \frac{1}{\sqrt{x^2+4}} ight) dx$$= \frac{9}{4} \cdot \frac{1}{2} 	an^{-1} \left( \frac{x}{2} ight) - \frac{1}{4} \left( -\frac{1}{x} ight) + \sinh^{-1} \left( \frac{x}{2} ight) + c$$= \frac{9}{8} 	an^{-1} \left( \frac{x}{2} ight) + \frac{1}{4x} + \sinh^{-1} \left( \frac{x}{2} ight) + c$.

$\int \frac{2x^2-1+x^2\sqrt{x^2+4}}{x^2(x^2+4)} dx =$

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