Integral Calculus Question 171
Question: $ \int_{{}}^{{}}{\frac{a\ dx}{b+ce^{x}}}= $
[MP PET 1988; BIT Ranchi 1979]
Options:
A) $ \frac{a}{b}\log ( \frac{e^{x}}{b+ce^{x}} )+c $
B) $ \frac{a}{b}\log ( \frac{b+ce^{x}}{e^{x}} )+c $
C) $ \frac{b}{a}\log ( \frac{e^{x}}{b+ce^{x}} )+c $
D) $ \frac{b}{a}\log ( \frac{b+ce^{x}}{e^{x}} )+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\frac{a,dx}{b+c,e^{x}}=\int_{{}}^{{}}{\frac{ae^{x}}{be^{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}{ \frac{c}{ct+b}-\frac{1}{t} }dt}} $ {By partial fraction}
$ =\frac{a}{b}\log ( \frac{e^{x}}{b+ce^{x}} )+c $ .
BETA
