Question

Resolve the function \( f(x)=\frac{x^{3}}{\left(x^{2}+1\right)(x+1)} \) into partial fractions.

Answer 1

f(x) = \(\frac{x^3}{(x^2+1)(x+1)}\) 

= \(\frac{x^3}{x^3+x^2+x+1}\) 

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

Now,

\(\frac{x^2+x+1}{(x^2+1)(x+1)}\) = \(\frac{A}{x+1}\) + \(\frac{Bx+c}{x^2+1}\)

⇒ x² + x + 1 = A(x²+1)+(Bx+c)(x+1)

Put x = -1, we get

2A = 1-1+1 = 1

⇒ A = \(\frac{1}{2}\)

Put x = 0, we get

A + C = 1

⇒ C = 1 - A

= 1 - \(\frac{1}{2}\)

= \(\frac{1}{2}\)

Put x = 1, we get

2A + 2B + 2C = 3

⇒ 2C = 3 - 2A - 2B 

= 3 - 1 - 1 

= 1

⇒ C = \(\frac{1}{2}\) 

Hence,

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

∴ f(x) = 1 - \(\frac{x^2+x+1}{(x^2+1)(x+1)}\)

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