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प्रश्न
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उत्तर
\[\text{ Let } I = \int\frac{\left( x - 1 \right)^2}{x^2 + 2x + 2} dx\]
\[ = \int\left( \frac{x^2 - 2x + 1}{x^2 + 2x + 2} \right) dx\]
\[\text{ Here }, \]
\[\text{ Thereforez }, \]
\[\frac{x^2 - 2x + 1}{x^2 + 2x + 2} = 1 - \frac{\left( 4x + 1 \right)}{x^2 + 2x + 2} . . . . . \left( 1 \right)\]
\[\text{ Let } 4x + 1 = A\frac{d}{dx} \left( x^2 + 2x + 2 \right) + B\]
\[4x + 1 = A \left( 2x + 2 \right) + B\]
\[4x + 1 = \left( 2A \right) x + 2A + B\]
\[ \text{ Equating Coefficients of like terms }\]
\[\text{ 2A = 4 }\]
\[A = 2\]
\[2A + B = 1\]
\[2 \times 2 + B = 1\]
\[B = - 3\]
\[\int\left( \frac{x^2 - 2x + 1}{x^2 + 2x + 2} \right) dx\]
\[ = \int dx - 2\int\frac{\left( 2x + 2 \right)}{x^2 + 2x + 2} dx + 3\int\frac{dx}{x^2 + 2x + 2}\]
\[ = \int dx - 2\int\frac{\left( 2x + 2 \right)}{x^2 + 2x + 2} dx + 3\int\frac{dx}{\left( x + 1 \right)^2 + 1^2}\]
\[ = x - 2 \text{ log } \left| x^2 + 2x + 2 \right| + \frac{3}{1} \tan^{- 1} \left( \frac{x + 1}{1} \right) + C\]
\[ = x - 2 \text { log } \left| x^2 + 2x + 2 \right| + 3 \tan^{- 1} \left( x + 1 \right) + C\]
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