Question

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

Answer 1

\[\int\left( x + 1 \right)e^x  . \text{ log } \left( x \text{ e}^x \right) dx\]
\[\text{ Let x e}^x = t\]
\[ \Rightarrow \left( x . e^x + 1 . e^x \right)dx = dt\]
\[ \therefore \int \left( x + 1 \right) e^x . \text{ log } \left( x \text{ e }^x \right) dx = \int 1_{II} . \text{ log }_I\left( t \right) dt\]
\[ = \text{ log  t }\int1\text{  dt } - \int\left\{ \frac{d}{dt}\left( \text{ log  t } \right) - \int1  \text{ dt }\right\}dt\]
\[ = \text{ log }\left( t \right) \times t - \int\frac{1}{t} \times \text{ t  dt }\]
\[ = \text{ t  log }\left( t \right) - t + C . . . (1)\]
\[\text{Substituting the value of t in eq}     \text{ (1) }\]
\[ \Rightarrow \int \left( x + 1 \right) e^x . \text{ log } \left( x \text{ e}^x \right) dx = \left( \text{ x e}^x \right) . \text{ log }\left( x \text{ e}^x \right) - \text{ x e }^x + C\]
\[ = \text{ x e}^x \left\{ \text{ log }\left( \text{ x e}^x \right) - 1 \right\} + C\]