Question

\[\int\frac{x + 1}{x \left( 1 + x e^x \right)} dx\]

Answer 1

We have,

\[I = \int\frac{x + 1}{x \left( 1 + x e^x \right)} dx\]
\[I = \int\frac{e^x \left( x + 1 \right)}{e^x x \left( 1 + x e^x \right)} dx\]
\[\text{Put }e^x = t\]
\[ \Rightarrow e^x \left( x + 1 \right)dx = dt\]
\[I = \int\frac{dt}{t \left( 1 + t \right)} . . . . . \left( 1 \right)\]
Let,
\[\frac{1}{t \left( 1 + t \right)} = \frac{A}{t} + \frac{B}{1 + t}\]
\[ \Rightarrow 1 = A\left( t + 1 \right) + Bt . . . . . \left( 2 \right)\]
\[\text{Putting t= 0 in (2), we obtain A = 1}\]
\[\text{Putting t = -1 in (2), we obtain B = -1}\]
\[I = \int\left( \frac{1}{t} - \frac{1}{1 + t} \right) dt\]
\[I = \log\left| t \right| - \log\left| t + 1 \right| + C\]
\[I = \log\left| \frac{t}{t + 1} \right| + C\]
\[I = \log\left| \frac{x e^x}{x e^x + 1} \right| + C\]