A) \[\frac{3}{2}x+\frac{35}{36}\log |9{{e}^{2x}}-4|+C\]
B) \[\frac{3}{2}x-\frac{35}{36}\log |9{{e}^{2x}}-4|+C\]
C) \[-\frac{3}{2}x+\frac{35}{36}\log |9{{e}^{2x}}-4|+C\]
D) \[-\frac{5}{2}x+\frac{35}{36}\log |9{{e}^{2x}}-4|+C\]
E) \[\frac{5}{2}x+\frac{35}{36}\log |9{{e}^{2x}}-4|+C\]
Correct Answer: C
Solution :
\[\int{\frac{4{{e}^{x}}+6{{e}^{-x}}}{9{{e}^{x}}-4{{e}^{-x}}}}dx\] \[=\int{\frac{4{{e}^{2x}}+6}{9{{e}^{2x}}-4}}dx\] \[=4\int{\frac{{{e}^{2x}}}{9{{e}^{2x}}-4}}dx+6\int{\frac{{{e}^{-2x}}}{9-4{{e}^{-2x}}}}dx\] Put\[{{t}_{1}}=9{{e}^{2x}}-4\]and\[{{t}_{2}}=9-4{{e}^{-2x}}\] \[d{{t}_{1}}=18{{e}^{2x}}dx\]and\[d{{t}_{2}}=8{{e}^{-2x}}dx\] \[=4\int{\frac{1}{{{t}_{1}}}.\frac{d{{t}_{1}}}{18}}+6\int{\frac{1}{{{t}_{2}}}.\frac{d{{t}_{2}}}{8}}\] \[=\frac{2}{9}\int{\frac{d{{t}_{1}}}{{{t}_{1}}}}+\frac{3}{4}\int{\frac{d{{t}_{2}}}{{{t}_{2}}}}\] \[=\frac{2}{9}\log {{t}_{1}}+\frac{3}{4}\log {{t}_{2}}+C\] \[=\frac{2}{9}\log |9{{e}^{2x}}-4|+\frac{3}{4}\log |9-4{{e}^{-2x}}|+C\] \[=\frac{2}{9}\log |9{{e}^{2x}}-4|+\frac{3}{4}\log |9{{e}^{2x}}-4|\] \[-\frac{3}{4}\log |{{e}^{2x}}|+C\] \[=\left( \frac{2}{9}+\frac{3}{4} \right)\log |9{{e}^{2x}}-4|-\frac{3}{4}(2x)+C\] \[(\because \log e=1)\] \[=-\frac{3}{2}x+\frac{35}{36}\log |9{{e}^{2x}}-4|+C\]
You need to login to perform this action.
You will be redirected in
3 sec
