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प्रश्न
\[\int_0^1 \frac{1}{1 + 2x + 2 x^2 + 2 x^3 + x^4}dx\]
योग
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उत्तर
\[\int_0^1 \frac{1}{1 + 2x + 2 x^2 + 2 x^3 + x^4}dx\]
\[ = \int_0^1 \frac{1}{\left( x^2 + 1 \right)^2 + 2x\left( x^2 + 1 \right)}dx\]
\[ = \int_0^1 \frac{1}{\left( x^2 + 1 \right)\left( x^2 + 1 + 2x \right)}dx\]
\[ = \int_0^1 \frac{1}{\left( x^2 + 1 \right) \left( x + 1 \right)^2}dx\]
\[ = \int_0^1 \frac{1}{\left( x^2 + 1 \right)^2 + 2x\left( x^2 + 1 \right)}dx\]
\[ = \int_0^1 \frac{1}{\left( x^2 + 1 \right)\left( x^2 + 1 + 2x \right)}dx\]
\[ = \int_0^1 \frac{1}{\left( x^2 + 1 \right) \left( x + 1 \right)^2}dx\]
Let
\[\frac{1}{\left( x + 1 \right)^2 \left( x^2 + 1 \right)} = \frac{A}{x + 1} + \frac{B}{\left( x + 1 \right)^2} + \frac{Cx + D}{x^2 + 1}\]
\[ \Rightarrow 1 = A\left( x + 1 \right)\left( x^2 + 1 \right) + B\left( x^2 + 1 \right) + \left( Cx + D \right) \left( x + 1 \right)^2\]
\[ \Rightarrow 1 = A\left( x + 1 \right)\left( x^2 + 1 \right) + B\left( x^2 + 1 \right) + \left( Cx + D \right) \left( x + 1 \right)^2\]
Putting x = −1, we have
1 = 2B
\[\Rightarrow B = \frac{1}{2}\]
Putting x = 0, we have
A + B + D = 1 .....(2)
Equating coefficient of x3 on both sides, we have
A + C = 0 .....(3)
Equating coefficient of x2 on both sides, we have
A + B + 2C + D = 0 .....(4)
⇒ 2C = −1 [Using (1)]
\[\Rightarrow C = - \frac{1}{2}\]
\[\therefore A = \frac{1}{2}\]
\[A = \frac{1}{2}, B = \frac{1}{2}\] and
\[C = - \frac{1}{2}\] in (4), we have
D = 0
D = 0
\[\therefore \int_0^1 \frac{1}{\left( x + 1 \right)^2 \left( x^2 + 1 \right)}dx\]
\[ = \int_0^1 \frac{\frac{1}{2}}{x + 1}dx + \int_0^1 \frac{\frac{1}{2}}{\left( x + 1 \right)^2}dx + \int_0^1 \frac{- \frac{1}{2}x}{x^2 + 1}\]
\[ = \left.\frac{1}{2} \log\left( x + 1 \right)\right|_0^1 + \left.\frac{1}{2} \times \left( - \frac{1}{x + 1} \right)\right|_0^1 - \frac{1}{4} \int_0^1 \frac{2x}{x^2 + 1}dx\]
\[ = \frac{1}{2}\left( \log2 - \log1 \right) - \frac{1}{2}\left( \frac{1}{2} - 1 \right) - \left.\frac{1}{4} \log\left( x^2 + 1 \right)\right|_0^1 \]
\[ = \frac{1}{2}\log2 + \frac{1}{4} - \frac{1}{4}\left( \log2 - \log1 \right) ................\left( \log1 = 0 \right)\]
\[ = \int_0^1 \frac{\frac{1}{2}}{x + 1}dx + \int_0^1 \frac{\frac{1}{2}}{\left( x + 1 \right)^2}dx + \int_0^1 \frac{- \frac{1}{2}x}{x^2 + 1}\]
\[ = \left.\frac{1}{2} \log\left( x + 1 \right)\right|_0^1 + \left.\frac{1}{2} \times \left( - \frac{1}{x + 1} \right)\right|_0^1 - \frac{1}{4} \int_0^1 \frac{2x}{x^2 + 1}dx\]
\[ = \frac{1}{2}\left( \log2 - \log1 \right) - \frac{1}{2}\left( \frac{1}{2} - 1 \right) - \left.\frac{1}{4} \log\left( x^2 + 1 \right)\right|_0^1 \]
\[ = \frac{1}{2}\log2 + \frac{1}{4} - \frac{1}{4}\left( \log2 - \log1 \right) ................\left( \log1 = 0 \right)\]
\[= \frac{1}{2}\log 2 + \frac{1}{4}\log e - \frac{1}{4}\log2\]
\[ = \frac{1}{4}\log 2 + \frac{1}{4}\log e\]
\[ = \frac{1}{4}\left( \log 2 + \log e \right)\]
\[ = \frac{1}{4}\log\left( 2e \right)\]
\[ = \frac{1}{4}\log 2 + \frac{1}{4}\log e\]
\[ = \frac{1}{4}\left( \log 2 + \log e \right)\]
\[ = \frac{1}{4}\log\left( 2e \right)\]
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