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प्रश्न
उत्तर
\[\text{ Let I } = \int e^x \left[ \frac{x - 1}{\left( x - 1 \right)^3} \right]dx\]
\[ = \int e^x \left[ \frac{x + 1 - 2}{\left( x + 1 \right)^3} \right]dx\]
\[ = \int e^x \left[ \frac{1}{\left( x - 1 \right)^2} - \frac{2}{\left( x + 1 \right)^3} \right]dx\]
\[\text{ Here}, f(x) = \frac{1}{\left( x + 1 \right)^2}\]
\[ \Rightarrow f'(x) = \frac{- 2}{\left( x + 1 \right)^2}\]
\[\text{ Put e}^x f(x) = t\]
\[\text{ let e}^x \frac{1}{\left( x + 1 \right)^2} = t\]
\[\text{ Diff both sides }\]
\[ e^x \frac{1}{\left( x + 1 \right)^2} + e^x \frac{\left( - 2 \right)}{\left( x + 1 \right)^3} = \frac{dt}{dx}\]
\[ \Rightarrow e^x \left[ \frac{1}{\left( x + 1 \right)^2} - \frac{2}{\left( x + 1 \right)^3} \right]dx = dt\]
\[ \therefore \int e^x \left[ \frac{1}{\left( x + 1 \right)^2} - \frac{2}{\left( x + 1 \right)^3} \right]dx = \int dt\]
\[ = t + C\]
\[ = \frac{e^x}{\left( x + 1 \right)^2} + C\]
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