Question

\[\int e^x \frac{1 + x}{\left( 2 + x \right)^2} \text{ dx }\]

Answer 1

\[\text{ Let I } = \int e^x \left[ \frac{1 + x}{\left( 2 + x \right)^2} \right]dx\]

\[ = \int e^x \left( \frac{2 + x - 1}{\left( 2 + x \right)^2} \right)dx\]

\[ = \int e^x \left[ \frac{1}{\left( 2 + x \right)} - \frac{1}{\left( 2 + x \right)^2} \right]dx\]

\[\text{ Here, } f(x) = \frac{1}{2 + x}\]

\[ \Rightarrow f'(x) = \frac{- 1}{\left( 2 + x \right)^2}\]

\[\text{ Put e }^x f(x) = t\]

\[ \Rightarrow e^x \frac{1}{x + 2} = t\]

\[\text{ Diff  both  sides  w . r . t x}\]

\[ e^x \frac{1}{x + 2} + e^x \frac{- 1}{\left( x + 2 \right)^2} = \frac{dt}{dx}\]

\[ \Rightarrow e^x \left[ \frac{1}{x + 2} - \frac{1}{\left( x + 2 \right)^2} \right]dx = dt\]

\[ \therefore \int e^x \left[ \frac{1}{\left( 2 + x \right)} - \frac{1}{\left( 2 + x \right)^2} \right]dx = \int dt\]

\[ \Rightarrow I = t + C\]

\[ = \frac{e^x}{2 + x} + C\]