\(\frac{x^{2}}{x^{2} + 1} + \frac{x + 4}{1 - x}\)
\(\frac{3}{1 - x} + \frac{2x + 1}{x^{2} + 1}\)
\(\frac{x^{2}}{1 - x} + \frac{x + 4}{x^{2} + 1}\)
\(\frac{3}{1 - x} + \frac{2x + 2}{x^{2} + 1}\)
Correct answer is B
\(\frac{x^{2} + x + 4}{(1 - x)(x^{2} + 1)} = \frac{A}{1 - x} + \frac{Bx + C}{x^{2} + 1}\)
= \(\frac{A(x^{2} + 1) + (Bx + C)(1 - x)}{(1 - x)(x^{2} + 1)}\)
\(\implies x^{2} + x + 4 = A(x^{2} + 1) + (Bx + C)(1 - x)\)
\(x^{2} + x + 4 = Ax^{2} + A + Bx - Bx^{2} - Cx + C\)
\(\implies (A - B)x^{2} = x^{2}; A - B = 1 ...... (i)\)
\((B - C)x = x; B - C = 1 ..... (ii)\)
\(A + C = 4 ...... (iii)\)
Solving the above simultaneous equations by any of the known methods, we get
\(A = 3, B = 2, C = 1\)
\(\therefore \frac{x^{2} + x + 4}{(1 - x)(x^{2} + 1)} = \frac{3}{1 - x} + \frac{2x + 1}{x^{2} + 1}\)