(x+1)2x3+x=Ax+Bx+Cx2+1x2+2x+1=A(x2+1)+(Bx+C)xPut x=0,A=1Comparing coefficients of x2 on both sides,A+B=1⇒B=0Comparing coefficients of x on both sides,C=2∴(x+1)2x3+x=1x+2x2+1

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$\frac{{(x + 1)}^{2}}{x^{3} + x} =$

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$\frac{1}{x} - \frac{2}{x^{2} + 1}$

$\frac{1}{x} + \frac{2}{x^{2} + 1}$

$\frac{1}{x} - \frac{1}{x^{2} + 1}$

$\frac{1}{x} + \frac{1}{x^{2} + 1}$

answer is A.

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Detailed Solution

$\frac{{(x + 1)}^{2}}{x^{3} + x} = \frac{A}{x} + \frac{B x + C}{x^{2} + 1}$

$x^{2} + 2 x + 1 = A (x^{2} + 1) + (B x + C) x$

Put $x = 0, A = 1$

Comparing coefficients of $x^{2}$ on both sides,

$A + B = 1 \Rightarrow B = 0$

Comparing coefficients of $x$ on both sides,

$C = 2$

$∴ \frac{{(x + 1)}^{2}}{x^{3} + x} = \frac{1}{x} + \frac{2}{x^{2} + 1}$

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