A) \[\ln \,\,\left| \frac{x{{e}^{x}}}{1+x{{e}^{x}}} \right|+\frac{1}{1+x{{e}^{x}}}+C\]
B) \[(1+x{{e}^{x}})+ln\left| \frac{x{{e}^{x}}}{1+x{{e}^{x}}} \right|+C\]
C) \[\frac{1}{1+x{{e}^{x}}}+ln\left| x{{e}^{x}}(1+x{{e}^{x}}) \right|+C\]
D) None of these
Correct Answer: A
Solution :
| [a] \[I=\int{\frac{(1+x)}{x{{(1+x{{e}^{x}})}^{2}}}dx=\int{\frac{(1+x){{e}^{x}}}{x{{e}^{x}}{{(1+x{{e}^{x}})}^{2}}}dx}}\] |
| Put \[x{{e}^{x}}=t\Rightarrow ({{e}^{x}}+x{{e}^{x}})dx=dt\] |
| \[I=\int{\frac{dt}{t{{(1+t)}^{2}}}}\] |
| Let \[\frac{1}{t{{(1+t)}^{2}}}=\frac{A}{t}+\frac{B}{1+t}+\frac{D}{{{(1+t)}^{2}}},\] we get |
| \[A=\frac{1}{{{(1+0)}^{2}}}=1,\,\,D=\frac{1}{-1}=-1\] |
| Equating coefficient of \[{{t}^{2}},0=A+B\Rightarrow B=-1\] |
| \[\therefore I=\int{\left[ \frac{1}{t}-\frac{1}{1+t}-\frac{1}{{{(1+t)}^{2}}} \right]dt}\] |
| \[=\,\,\,\,ln\left| t \right|-ln\left| 1+t \right|+\frac{1}{1+t}+C\] |
| \[=\,\,ln\left| \frac{t}{1+t} \right|+\frac{1}{1+t}+C\] |
| \[=\,\,\,ln\left| \frac{x{{e}^{x}}}{1+x{{e}^{x}}} \right|+\frac{1}{1+x{{e}^{x}}}+C\] |
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