If a^x e^ 2x b^x c^x dx = 1 k ( a^x e^ 2x b^x c^x ) + l then k = - Vidyadip

MCQ

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 $k =$

A
$log\, b + log \,c -log\, a -2$
B
$log\, (e^2 \,abc)$
✓
$log\, a -log\, b -log\, c + 2$
D
$2\, log\, a + log\, b -log\, c$

✓

Answer

Correct option: C.

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

c
$\int\left(\frac{\mathrm{ae}^{2}}{\mathrm{bc}}\right)^{\mathrm{x}} \mathrm{dx}$

$\frac{{{{\left( {\frac{{{\rm{a}}{{\rm{e}}^2}}}{{{\rm{bc}}}}} \right)}^{\rm{x}}}}}{{\ln \frac{{{\rm{a}}{{\rm{e}}^2}}}{{{\rm{bc}}}}}} + l$

$\therefore {\rm{k}} = \ln \frac{{{\rm{a}}{{\rm{e}}^2}}}{{{\rm{bc}}}}$

$ = \ln {\rm{a}} + 2 - \ln {\rm{b}} - \ln {\rm{c}}$

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