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∫ X 2 − 2 X 5 − X Dx - Mathematics

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Sum
\[\int\frac{x^2 - 2}{x^5 - x} \text{ dx}\]
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Solution

\[\text{We have}, \]
\[I = \int\left( \frac{x^2 - 2}{x^5 - x} \right) dx\]
\[ = \int\frac{\left( x^2 - 2 \right)}{x \left( x^4 - 1 \right)}dx\]
\[ = \int\frac{x \left( x^2 - 2 \right)}{x^2 \left( x^2 - 1 \right) \left( x^2 + 1 \right)}dx\]
\[\text{ Putting x^2 = t}\]
\[ \Rightarrow 2x\ dx = dt\]
\[ \Rightarrow x\ dx = \frac{dt}{2}\]
\[ \therefore I = \frac{1}{2}\int\frac{\left( t - 2 \right)}{t \left( t - 1 \right) \left( t + 1 \right)}\text{  dt }\]
\[\text{ Let }  \frac{t - 2}{t \left( t - 1 \right) \left( t + 1 \right)} = \frac{A}{t} + \frac{B}{t - 1} + \frac{C}{t + 1}\]
\[ \Rightarrow \frac{t - 2}{t \left( t - 1 \right) \left( t + 1 \right)} = \frac{A \left( t - 1 \right) \left( t + 1 \right) + Bt \left( t + 1 \right) + Ct \cdot \left( t - 1 \right)}{t \left( t - 1 \right) \left( t + 1 \right)}\]
\[ \Rightarrow t - 2 = A \left( t - 1 \right) \left( t + 1 \right) + B t \left( t + 1 \right) + C t \left( t - 1 \right)\]
\[\text{ Putting t = 1}\]
\[ \therefore 1 - 2 = B \times 2\]
\[ \Rightarrow B = - \frac{1}{2}\]
\[\text{ Putting t = 0}\]
\[ \therefore - 2 = A \left( - 1 \right)\]
\[ \Rightarrow A = 2\]
\[\text{ Putting t = - 1}\]
\[ \therefore - 3 = C \left( - 1 \right) \left( - 2 \right)\]
\[ \Rightarrow C = - \frac{3}{2}\]
\[ \therefore I = \frac{2}{2}\int\frac{dt}{t} - \frac{1}{2 \times 2}\int\frac{dt}{t - 1} - \frac{3}{2 \times 2}\int\frac{d}{t + 1}\]
\[ = \text{ log} \left| t \right| - \frac{1}{4} \text{ log }\left| t - 1 \right| - \frac{3}{4} \text{ log} \left| t + 1 \right| + C\]
\[ = \text{ log }\left| x^2 \right| - \frac{1}{4} \text{ log }\left| x^2 - 1 \right| - \frac{3}{4} \text{ log} \left| x^2 + 1 \right| + C\]
\[ = 2 \text{ log } \left| x \right| - \frac{1}{4} \text{ log} \left| x^2 - 1 \right| - \frac{3}{4} \text{ log} \left| x^2 + 1 \right| + C\]

Concept: Indefinite Integral Problems
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APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Revision Excercise | Q 126 | Page 205

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