_ ^ a ;dx b + c e^x = - Vidyadip

MCQ

$\int_{}^{} {\frac{{a\;dx}}{{b + c{e^x}}}} = $

✓
$\frac{a}{b}\log \left( {\frac{{{e^x}}}{{b + c{e^x}}}} \right) + c$
B
$\frac{a}{b}\log \left( {\frac{{b + c{e^x}}}{{{e^x}}}} \right) + c$
C
$\frac{b}{a}\log \left( {\frac{{{e^x}}}{{b + c{e^x}}}} \right) + c$
D
$\frac{b}{a}\log \left( {\frac{{b + c{e^x}}}{{{e^x}}}} \right) + c$

✓

Answer

Correct option: A.

$\frac{a}{b}\log \left( {\frac{{{e^x}}}{{b + c{e^x}}}} \right) + c$

a
(a)$\int_{}^{} {\frac{{a\,dx}}{{b + c\,{e^x}}} = \int_{}^{} {\frac{{a{e^x}}}{{b{e^x} + c\,{e^{2x}}}}\,dx} } $
Now put ${e^x} = t,$ then it reduces to
$a\int_{}^{} {\frac{{dt}}{{t(ct + b)}} = a\int_{}^{} { - \frac{1}{b}\left\{ {\frac{c}{{ct + b}} - \frac{1}{t}} \right\}dt} } $ {By partial fraction}
$ = \frac{a}{b}\log \left( {\frac{{{e^x}}}{{b + c{e^x}}}} \right) + c$.

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