Integral Calculus Question 403
Question: If $ \int_{{}}^{{}}{\frac{4e^{x}+6{e^{-x}}}{9e^{x}-4{e^{-x}}}dx=Ax+B\log (9e^{2x}-4)}+C $ , then A, B and C are
[IIT 1990]
Options:
A) $ A=\frac{3}{2},\ B=\frac{36}{35},\ C=\frac{3}{2}\log 3+ $ constant
B) $ A=\frac{3}{2},\ B=\frac{35}{36},\ C=\frac{3}{2}\log 3+ $ constant
C) $ A=-\frac{3}{2},\ B=-\frac{35}{36},\ C=-\frac{3}{2}\log 3+ $ constant
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
$ I=\int_{{}}^{{}}{\frac{4e^{x}+6{e^{-x}}}{9e^{2x}-4{e^{-x}}},dx}=\frac{4}{9}\int_{{}}^{{}}{\frac{9e^{2x}dx}{9e^{2x}-4}+6}\int_{{}}^{{}}{\frac{dx}{9e^{2x}-4}} $
$ \therefore ,\int_{{}}^{{}}{\frac{dx}{9e^{2x}-4}}=\frac{1}{8}\log (9e^{2x}-4)-\frac{1}{4}\log 3-\frac{1}{4}x+ $ const.
$ \therefore ,I=\frac{35}{36}\log (9e^{2x}-4)-\frac{3}{2}x-\frac{3}{2}\log 3+ $ const.
Comparing with the given integral, we get
$ A=-\frac{3}{2}, $ $ B=\frac{35}{36}, $ $ C=-\frac{3}{2}\log 3+ $ const.
BETA
