lets see if I am understanding this any better: integral (2e^u)+(6/u)+(ln2)dx = e^2u + 6(lnu) + 1/2 + C
Follow Math Help Forum on Facebook and Google+
Originally Posted by becky lets see if I am understanding this any better: integral (2e^u)+(6/u)+(ln2)dx = e^2u + 6(lnu) + 1/2 + C $\displaystyle \int 2e^x + \frac{6}{x} + \ln 2 dx = \int 2e^x dx + \int \frac{6}{x} dx + \int \ln 2 dx$ $\displaystyle \int 2e^x = 2e^x +C$ $\displaystyle \int \frac{6}{x} dx = 6\ln |x|+C$ $\displaystyle \int \ln 2 dx = x\ln 2 +C$ Thus, the total sum is the answer.
Last edited by ThePerfectHacker; Jun 4th 2007 at 07:17 PM.
how come 2e^u isn't supposed to be e^2u or e^2^u?
Originally Posted by becky how come 2e^u isn't supposed to be e^2u or e^2^u? Because if it were $\displaystyle e^{2u}$ then the derivative would be $\displaystyle 2e^{2u}$. Which is wrong.
View Tag Cloud