Question #174633

Evaluate the integral of e^(2lnx) dx


1
Expert's answer
2021-03-31T11:55:00-0400

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

Answer: x33+C\frac{{{x^3}}}{3} + C


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