Q64 / 160 · TS EAMCET 2023 (12 May Shift 2) · Mathematics
If $\frac{x+1}{\left(x^2+1\right)(x-1)^2}=\frac{A x+B}{x^2+1}+\frac{C}{x-1}+\frac{D}{(x-1)^2}$, then $\mathrm{A}+\mathrm{B}+\mathrm{C}+\mathrm{D}=$
- A$-\frac{1}{2}$
- Correct answerB$\frac{1}{2}$
- C$1$
- D$\frac{3}{2}$
Solution
$$
\begin{aligned}
& \quad \frac{x+1}{\left(x^2+1\right)(x-1)^2}=\frac{A x+B}{x^2+1}+\frac{C}{x-1}+\frac{D}{(x-1)^2} \\
& (x+1)=(A x+B)(x-1)^2+C(x-1)\left(x^2+1\right)+D\left(x^2+1\right) \\
& \begin{array}{r}
\text { Put } x=1 \\
2=2 D \Rightarrow D=1 \\
x+1=(A x+B)\left(x^2+1-2 x\right) \\
x+1=(A+C) x^3+(-2 A+B-C+D) x^2 \\
+(A-2 B+C) x+(B-C+D)
\end{array}
\end{aligned}
$$
Comparing coefficients
$$
\begin{aligned}
& A+C=0 \Rightarrow C=-A \\
& -2 A+B-C+D=0 \\
& \Rightarrow-A+B+1=0 \Rightarrow A-2 B+C=1 \\
& \Rightarrow-2 B=1 \Rightarrow B=\frac{-1}{2}
\end{aligned}
$$
and $B-C+D=1$
$$
\begin{aligned}
& \Rightarrow B-C=0 \Rightarrow C=\frac{-1}{2}, A=\frac{1}{2} \\
& \therefore \quad A+B+C+D=\frac{1}{2}-\frac{1}{2}-\frac{1}{2}+1=\frac{1}{2} .
\end{aligned}
$$
Hence, the correct answer is option: 'B'.