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Integral I=∫e3log⁡x−3x4e2log⁡x+x2 dxI = \displaystyle\int \dfrac{e^{3\log x} - 3x^4}{e^{2\log x} + x^2}\, dx (where x>0x>0) is equal to

Solution

✅ Correct Option: 4

Since e3log⁡x=x3e^{3\log x}=x^3 and e2log⁡x=x2e^{2\log x}=x^2, the integrand becomes x3−3x42x2=x2−3x22\dfrac{x^3-3x^4}{2x^2}=\dfrac{x}{2}-\dfrac{3x^2}{2}.

Integrating gives x24−x32+C\dfrac{x^2}{4}-\dfrac{x^3}{2}+C.

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