Question

Partial fraction of $\frac{x^2+1}{x^3+1}$ will be of form:

• A. $\frac{A}{x+1}-Bx+\frac{C}{x^2-x+1}$
• B. $\frac{A}{x+1}+Bx+\frac{C}{x^2-x+1}$
• C. $\frac{A}{x+1}+Bx-\frac{C}{x^2-x+1}$
• D. None of these

Answer 1

The correct option is A $\frac{A}{x+1}-Bx+\frac{C}{x^2-x+1}$

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

$\therefore y=\frac{A}{x+1}+\frac{Bx+C}{x^2-x+1}$

$A(x^2-x+1)+(Bx+C)(x+1)=x^2+1$

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

$\therefore A+B=1,A=B+C,A+C=1$

$\therefore A=\frac{2}{3},B=\frac{1}{3}=C$

$\therefore y=\frac{1}{3}[\frac{2}{x+1}+\frac{x+1}{x^2-x+1}]$