Question

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

Answer 1

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

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

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

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

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

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

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

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

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

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

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

\[ \Rightarrow I = t + C\]

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