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integrate (e^5x)/[e^10x -e^5x]
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\[\int\limits \frac{ e ^{5x} }{e ^{10x}-e ^{5x} }dx\] put \[e ^{5x}=u,e ^{5x}*5 dx=du,dx=\frac{ 1 }{5 }du\] \[also~\left( e ^{5x} \right)^{2}=u ^{2},e ^{10x}=u ^{2}\] substitute and integrate after making partial fractions.
hmm.. so then I get (1/5)int[u/(u(u-1))] --> (1/5)int(u-1)^1 du = ln|e^5x -1| +C?
no \[\frac{ 1 }{5 }\int\limits \frac{ du }{ u \left( u-1 \right) }\]
Oh, thanks that helps!
yw
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