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प्रश्न
\[\int\frac{x^3}{\sqrt{x^8 + 4}} \text{ dx }\]
बेरीज
उत्तर
\[\text{ Let I }= \int\frac{x^3}{\sqrt{x^8 + 2^2}}dx\]
\[ = \int\frac{x^3}{\sqrt{\left( x^4 \right)^2 + 2^2}}dx\]
\[\text{ Putting x}^4 = t\]
\[ \Rightarrow 4 x^3 \text{ dx }= dt\]
\[ \Rightarrow x^3 \cdot dx = \frac{dt}{4}\]
\[ \therefore I = \frac{1}{4}\int\frac{1}{\sqrt{t^2 + 2^2}}dt\]
\[ = \frac{1}{4} \text{ ln} \left| t + \sqrt{t^2 + 4} \right| + C\]
\[ = \frac{1}{4} \text{ ln }\left| x^4 + \sqrt{x^8 + 4} \right| + C ...........\left[ \because t = x^4 \right]\]
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