Question

If $\int \frac{4e^x+6e^{-x}}{9e^x-4e^{-x}}dx=Ax+bln(9e^{2x}-4)+C;$ then; value of A, B, & C are

• A. $A=-\frac{3}{2},B=\frac{35}{36},C\epsilon R$
• B. $A=\frac{3}{2},B=\frac{-35}{36},C\epsilon R$
• C. $A=-\frac{3}{2},B=\frac{35}{36},C>0$
• D. None of these

Answer 1

The correct option is A $A=-\frac{3}{2},B=\frac{35}{36},C\epsilon R$

$I=\int \frac{{4e}^{2x}+6}{{9e}^{2x}-4}dx,$ put ${9e}^{2x}-4=z.$
Then $I=\int \frac{1}{z}\{4.\frac{z+4}{9}+6\}\frac{1}{18}\frac{dz}{\frac{z+4}{9}}$
$=\int \frac{1}{z(z+4)}\{\frac{2z+8}{9}+6\}dz$
$=\frac{1}{9}\int \frac{2z+35}{z(z+4)}dz=\frac{1}{9}\int \frac{2(z+4)+27}{z(z+4)}dz$
$=\frac{2}{9}\int \frac{dz}{z}+\frac{3}{4}\int (\frac{1}{z}-\frac{1}{z+4})dz$
$=(\frac{2}{9}+\frac{3}{4})\log z-\frac{3}{4}\log (z+4)+C$
$=\frac{35}{36}\log ({9e}^{2x}-4)=\frac{3}{4}\log \left(e^{2x}\right)+C$
$=-\frac{3}{2}\times +\frac{35}{36}\log ({9e}^{2x}-4)-\frac{3}{2}\log 3+C$