Question

If $\int \frac{3e^x+5e^{-x}}{4e^x-5e^{-x}}dx=Ax+B\ln |4e^{2x}-5|+K$ for constant of integration $K,$ then

• A. $A=-1$ and $B=-\frac{7}{8}$
• B. $A=1$ and $B=\frac{7}{8}$
• C. $A=-\frac{1}{8}$ and $B=\frac{7}{8}$
• D. $A=-1$ and $B=\frac{7}{8}$

Answer 1

The correct option is D $A=-1$ and $B=\frac{7}{8}$

Let
$I=\int \frac{3e^x+5e^{-x}}{4e^x-5e^{-x}}dx=\int \frac{3e^{2x}+5}{4e^{2x}-5}dx$

Let $e^{2x}=t$
$\Rightarrow 2e^{2x}dx=dt\Rightarrow dx=\frac{dt}{2t}$

$I=\int \frac{3t+5}{2t(4t-5)}dt$
Using partial fraction,
$I=\frac{1}{2}\left(\int \frac{-1}{t}dt+\int \frac{7}{(4t-5)}dt\right)=\frac{1}{2}(-\ln t+\frac{7}{4}\ln |4t-5|)+K=-\frac{1}{2}\ln e^{2x}+\frac{7}{8}\ln |4e^{2x}-5|+K=-x+\frac{7}{8}\ln |4e^{2x}-5|+K$
$\therefore A=-1$ and $B=\frac{7}{8}$