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\int\limits_{1}^{e} (2x+e)/(x^2+ex)
i mean \[\int\limits_{1}^{e} (2x+e)/(x^2+ex)\]
oh there is meant to be a" dx "at the end as well
guess you want a u - sub for this one put \[u=x^2+ex\] \[du=(2x+e)dx\] \[u(1)=1+e\] \[u(e)= e^2+e\] so get \[\int_{1+e}^{e^2+e} \frac{du}{u}\]
gives \[\ln(u)|_{1+e}^{e^2+e}\] \[=\ln(e^2+e)-\ln(1+e)\]
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\[=\ln(\frac{e^2+e}{1+e})=\ln(e)=1\]\]
thank math whiz
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