Question

If $\int \frac{4e^x+6e^{-x}}{9e^x-4e^{-x}}dx=Ax+Blog(9e^{2x}-4)+C$, then A, B and C are

• A. $A=\frac{3}{2},B=\frac{36}{35},C=\frac{3}{2}log3+constant$
• B. $A=\frac{3}{2},B=\frac{35}{36},C=\frac{3}{2}log3+constant$
• C. $A=-\frac{3}{2},B=\frac{35}{36},C=-\frac{3}{2}log3+constant$
• D. None of these

Answer 1

The correct option is C $A=-\frac{3}{2},B=\frac{35}{36},C=-\frac{3}{2}log3+constant$

$I=\int \frac{4e^x+6e^{-x}}{9e^x-4e^{-x}}dx=\frac{4}{9}\int \frac{9e^{2x}dx}{9e^{2x}-4}+6\int \frac{dx}{9e^{2x}-4}\therefore \int \frac{dx}{9e^{2x}-4}=\frac{1}{8}log(9e^{2x}-4)-\frac{1}{4}log3-\frac{1}{4}x+const.\therefore I=\frac{35}{36}log(9e^{2x}-4)-\frac{3}{2}x-\frac{3}{2}log3+const.$
Comparing with the given integral, we get
$A=-\frac{3}{2},B=\frac{35}{36},C=-\frac{3}{2}log3+const.$