Question

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

• A. $A=\frac{19}{36},B=\frac{35}{36},C=0$
• B. $A=\frac{-19}{36},B=\frac{35}{36},C=0$
• C. $A=\frac{-19}{36},B=\frac{-35}{36},C\in R$
• D. $A=\frac{35}{36},B=\frac{-19}{36},C\in R$

Answer 1

The correct option is D $A=\frac{35}{36},B=\frac{-19}{36},C\in R$

$\int \frac{4e^x+6e^{-x}}{9e^x+4e^{-x}}dx=Ax+B\log \left(9e^x+4^{-x}\right)+C4e^x+6e^{-x}=A\left(9e^x+4e^{-x}\right)+B\left(9e^x+4^{-x}\right)rA+9B=49A+9B=44A-4B=62A-2B=39A+9B=4\underset{–}{9A-9B=\frac{27}{2}}18A=4+\frac{27}{2}=\frac{35}{2}A=\frac{15}{36}2A-3=2B\frac{35}{18}-3=2B\frac{-19}{36}=B\int \frac{A\left(9e^x+4e^{-x}\right)+B\left(9e^x-4e^{-x}\right)}{9e^x+4e^{-x}}dx=Ax+B\log \left(9e^x+4e^{-x}\right)+CA=\frac{35}{36}B=\frac{-19}{36}C\in R$

Option $D$ is correct answer.