A) \[\log ({{e}^{x}}-1)-\log ({{e}^{x}}+2)+c\]
B) \[\frac{1}{2}\log ({{e}^{x}}-1)-\frac{1}{3}\log ({{e}^{x}}+2)+c\]
C) \[\frac{1}{3}\log ({{e}^{x}}-1)-\frac{1}{3}\log ({{e}^{x}}+2)+c\]
D) \[\frac{1}{3}\log ({{e}^{x}}-1)+\frac{1}{3}\log ({{e}^{x}}+2)+c\]
Correct Answer: C
Solution :
\[\int_{{}}^{{}}{\frac{{{e}^{x}}dx}{{{e}^{2x}}+{{e}^{x}}-2}}=\int_{{}}^{{}}{\frac{dt}{{{t}^{2}}+t-2}}\] \[\left\{ \because \,\,\,{{e}^{x}}=t\Rightarrow {{e}^{x}}dx=dt \right\}\] \[=\int_{{}}^{{}}{\frac{dt}{(t+2)(t-1)}}=\,\int_{{}}^{{}}{\frac{1}{3}\left[ \frac{1}{t-1}-\frac{1}{t+2} \right]}\,dt\] \[=\frac{1}{3}\log ({{e}^{x}}-1)-\frac{1}{3}\log ({{e}^{x}}+2)+c.\]
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