Advertisements
Advertisements
Question
Advertisements
Solution
We have,
\[I = \int\frac{x^2 dx}{\left( x - 1 \right) \left( x + 1 \right)^2}\]
\[\text{Let }\frac{x^2}{\left( x - 1 \right) \left( x + 1 \right)^2} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{\left( x + 1 \right)^2}\]
\[ \Rightarrow \frac{x^2}{\left( x - 1 \right) \left( x + 1 \right)^2} = \frac{A \left( x + 1 \right)^2 + B \left( x + 1 \right) \left( x - 1 \right) + C \left( x - 1 \right)}{\left( x + 1 \right)^2 \left( x - 1 \right)}\]
\[ \Rightarrow x^2 = A \left( x^2 + 2x + 1 \right) + B \left( x^2 - 1 \right) + C \left( x - 1 \right)\]
\[ \Rightarrow x^2 = \left( A + B \right) x^2 + x \left( 2A + C \right) + \left( A - B - C \right)\]
\[\text{Equating coefficients of like terms}\]
\[A + B = 1 ...................(1)\]
\[2A + C = 0 ....................(2)\]
\[A - B - C = 0 .......................(3)\]
\[\text{Solving (1), (2) and (3), we get}\]
\[A = \frac{1}{4}, B = \frac{3}{4}\text{ and }C = - \frac{1}{2}\]
\[ \therefore \frac{x^2}{\left( x - 1 \right) \left( x + 1 \right)^2} = \frac{1}{4 \left( x - 1 \right)} + \frac{3}{4 \left( x + 1 \right)} - \frac{1}{2 \left( x + 1 \right)^2}\]
\[ \Rightarrow I = \frac{1}{4}\int\frac{dx}{x - 1} + \frac{3}{4}\int\frac{dx}{x + 1} - \frac{1}{2}\int\frac{dx}{\left( x + 1 \right)^2}\]
\[ = \frac{1}{4} \log \left| x - 1 \right| + \frac{3}{4} \log \left| x + 1 \right| - \frac{1}{2} \times \frac{- 1}{x + 1} + C\]
\[ = \frac{1}{4}\log \left| x - 1 \right| + \frac{3}{4} \log \left| x + 1 \right| + \frac{1}{2 \left( x + 1 \right)} + C\]
APPEARS IN
RELATED QUESTIONS
Integrate the following integrals:
` ∫ e^{m sin ^-1 x}/ \sqrt{1-x^2} ` dx
