If int {frac{{{a^x}{e^{2x}}}}{{{b^x}{c^x}}}dx | vedclass

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(C) Given integral is $\int {\frac{{{a^x}{e^{2x}}}}{{{b^x}{c^x}}}} dx = \int {\left( {\frac{{a{e^2}}}{{bc}}} \right)^x} dx$. Using the formula $\int {

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$log\, a - log\, b - log\, c + 2$

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(C) Given integral is $\int {\frac{{{a^x}{e^{2x}}}}{{{b^x}{c^x}}}} dx = \int {\left( {\frac{{a{e^2}}}{{bc}}} \right)^x} dx$. Using the formula $\int {m^x} dx = \frac{{m^x}}{{\ln m}} + C$,where $m = \frac{{a{e^2}}}{{bc}}$,we get: $\int {\left( {\frac{{a{e^2}}}{{bc}}} \right)^x} dx = \frac{{{{\left( {\frac{{a{e^2}}}{{bc}}} \right)}^x}}}{{\ln \left( {\frac{{a{e^2}}}{{bc}}} \right)}} + l$. Comparing this with the given form $\frac{1}{k}\left( {\frac{{{a^x}{e^{2x}}}}{{{b^x}{c^x}}}} \right) + l$,we identify $k = \ln \left( {\frac{{a{e^2}}}{{bc}}} \right)$. Using logarithmic properties,$k = \ln a + \ln(e^2) - \ln b - \ln c = \ln a + 2 - \ln b - \ln c$. Thus,$k = \ln a - \ln b - \ln c + 2$.

If $\int {\frac{{{a^x}{e^{2x}}}}{{{b^x}{c^x}}}dx = \frac{1}{k}\left( {\frac{{{a^x}{e^{2x}}}}{{{b^x}{c^x}}}} \right)} + l$,then $k =$

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