my problem is to find the arc length of [tex]y=e^x[/tex] between 0 and 1(adsbygoogle = window.adsbygoogle || []).push({});

what i've got is [tex]\int_{0}^{1} \sqrt{1+(e^x)^2}dx[/tex] which i then substitute [tex]u=e^x[/tex] giving [tex]\int_{1}^{e} \frac{\sqrt{1+u^2}}{u^2}u du[/tex] which i then sub in v=[tex]\sqrt{1+u^2}[/tex] giving [tex]\int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v^2}{(v^2-1)^2} dv = \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v}{(v^2-1)} dv = \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{v}{(v-1)(v+1)} dv[/tex] which then needs to be integrated using partial fractions, and this is where i run into problems

i get [tex]\frac{1}{2} \int_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{1}{v-1}+\frac{1}{v+1}dv[/tex] which results in my getting an answer of 1.

the book on the other hand gets [tex]\int_{\sqrt{2}}^{\sqrt{1+e^2}}1+ \frac{\frac{1}{2}}{v-1}-\frac{\frac{1}{2}}{v+1}dv[/tex] but i cant see how they got the 1, or the negative sign on the second fraction

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# Homework Help: Integral problem.(arc length)

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