Evaluate the integral of e^(2lnx) dx
∫e2lnxdx=∫elnx2dx=∫x2dx=x33+C\int {{e^{2\ln x}}dx} = \int {{e^{\ln {x^2}}}dx} = \int {{x^2}} dx = \frac{{{x^3}}}{3} + C∫e2lnxdx=∫elnx2dx=∫x2dx=3x3+C
Answer: x33+C\frac{{{x^3}}}{3} + C3x3+C
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