If int {frac{{{a^x}{e^{3x}}}}{{{b^x}{c^x}}}} | vedclass

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Question

(B) We are given the integral $\int {\frac{{{a^x}{e^{3x}}}}{{{b^x}{c^x}}}} dx$. This can be rewritten as $\int {\left( {\frac{{a{e^3}}}{{bc}}} \right)

Vedclass · Accepted Answer

$\log a + 3 - \log b - \log c$

Vedclass · Answer

(B) We are given the integral $\int {\frac{{{a^x}{e^{3x}}}}{{{b^x}{c^x}}}} dx$. This can be rewritten as $\int {\left( {\frac{{a{e^3}}}{{bc}}} \right)^x} dx$. Using the standard integration formula $\int {A^x} dx = \frac{{A^x}}{{\log A}} + K$,where $A = \frac{{a{e^3}}}{{bc}}$,we get: $\int {\left( {\frac{{a{e^3}}}{{bc}}} \right)^x} dx = \frac{{{{\left( {\frac{{a{e^3}}}{{bc}}} \right)}^x}}}{{\log \left( {\frac{{a{e^3}}}{{bc}}} \right)}} + K$. Comparing this with the given expression $\frac{1}{P}\left( {\frac{{{a^x}{e^{3x}}}}{{{b^x}{c^x}}}} \right) + K$,we identify $P = \log \left( {\frac{{a{e^3}}}{{bc}}} \right)$. Using logarithmic properties,$P = \log (a{e^3}) - \log (bc) = \log a + \log ({e^3}) - \log (bc) = \log a + 3 - \log b - \log c$.

If $\int {\frac{{{a^x}{e^{3x}}}}{{{b^x}{c^x}}}} dx = \frac{1}{P}\left( {\frac{{{a^x}{e^{3x}}}}{{{b^x}{c^x}}}} \right) + K$; then $P =$ ?

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