Question

\[\int\frac{3}{\left( 1 - x \right) \left( 1 + x^2 \right)} dx\]

Answer 1

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