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∫ 1 1 + X + X 2 + X 3 Dx - Mathematics

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प्रश्न

\[\int\frac{1}{1 + x + x^2 + x^3} \text{ dx }\]
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उत्तर

\[\text{We have}, \]
\[I = \int\frac{dx}{1 + x + x^2 + x^3}\]
\[ = \int\frac{dx}{\left( 1 + x \right) + x^2 \left( 1 + x \right)}\]
\[ = \int \frac{dx}{\left( 1 + x \right) \left( 1 + x^2 \right)}\]
\[\text{ Let }\frac{1}{\left( x + 1 \right) \left( 1 + x^2 \right)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 1}\]
\[ \Rightarrow \frac{1}{\left( x + 1 \right) \left( x^2 + 1 \right)} = \frac{A \left( x^2 + 1 \right) + \left( Bx + C \right) \left( x + 1 \right)}{\left( x + 1 \right) \left( x^2 + 1 \right)}\]
\[ \Rightarrow 1 = A \left( x^2 + 1 \right) + B x^2 + Bx + Cx + C\]
\[ \Rightarrow 1 = \left( A + B \right) x^2 + \left( B + C \right) x + \left( A + C \right)\]
\[\text{Equating Coefficient of like terms}\]
\[A + B = 0 . . . . . \left( 1 \right)\]
\[B + C = 0 . . . . . \left( 2 \right)\]
\[A + C = 1 . . . . . \left( 3 \right)\]
\[\text{Solving} \left( 1 \right), \left( 2 \right) \text{ and }\left( 3 \right), \text{we get}, \]
\[A = \frac{1}{2}\]
\[B = - \frac{1}{2}\]
\[C = \frac{1}{2}\]
\[ \therefore \frac{1}{\left( x + 1 \right) \left( x^2 + 1 \right)} = \frac{1}{2 \left( x + 1 \right)} + \frac{- \frac{x}{2} + \frac{1}{2}}{x^2 + 1}\]
\[ \Rightarrow \frac{1}{\left( x + 1 \right) \left( x^2 + 1 \right)} = \frac{1}{2 \left( x + 1 \right)} - \frac{1}{2} \left( \frac{x}{x^2 + 1} \right) + \frac{1}{2 \left( x^2 + 1 \right)}\]
\[ \therefore I = \frac{1}{2}\int\frac{dx}{x + 1} - \frac{1}{2}\int\frac{x dx}{\left( x^2 + 1 \right)} + \frac{1}{2}\int\frac{dx}{x^2 + 1}\]
\[\text{ Putting x}^2 + 1 = t\]
\[ \Rightarrow 2x\ dx\ = dt\]
\[ \Rightarrow x\ dx\ = \frac{dt}{2}\]
\[ \therefore I = \frac{1}{2}\int\frac{dx}{x + 1} - \frac{1}{4}\int\frac{dt}{t} + \frac{1}{2}\int\frac{dx}{x^2 + 1}\]
\[ = \frac{1}{2} \text{ log }\left| x + 1 \right| - \frac{1}{4} \text{ log }\left| t \right| + \frac{1}{2} \text{ tan}^{- 1} x + C\]
\[ = \frac{1}{2} \text{ log }\left| x + 1 \right| - \frac{1}{4} \text{ log }\left| x^2 + 1 \right| + \frac{1}{2} \text{ tan}^{- 1} \left( x \right) + C\]
\[ = \frac{1}{2} \text{ log } \left| x + 1 \right| - \frac{1}{2} \text{ log } \left( \sqrt{x^2 + 1} \right) + \frac{1}{2} \text{ tan}^{- 1} \left( x \right) + C\]
\[ = \frac{1}{2} \text{ log } \left( \frac{\left| x + 1 \right|}{\sqrt{x^2 + 1}} \right) + \frac{1}{2} \text{ tan}^{- 1} \left( x \right) + C\]

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अध्याय 19: Indefinite Integrals - Revision Excercise [पृष्ठ २०५]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Revision Excercise | Q 124 | पृष्ठ २०५

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