Question

Please help! :)
Write the partial fraction decomposition of (3x^2 + 4)/(x^2 + 1)^2

Answer 1

\[\frac{ 3x^2 + 4 }{ (x^2 + 1)^2 } = \frac{ Ax + B }{ x^2 +1 } + \frac{ Cx + D }{ (x^2 + 1)^2 }\]

Answer 2

You need to add them (which will require identical denominators, obtained by manipulation) and solve for A, B, C and D.

Answer 3

I'm not sure what to do.

Answer 4

To make denominators same:

\[\frac{ Ax + B }{ x^2 + 1 } \times \frac{ x^2 + 1 }{ x^2 + 1 } + \frac{ Cx + D }{ (x^2+1)^2 }\]
\[= \frac{ Ax^3 + Ax + Bx^2 + B + Cx + D }{ (x^2+1)^2 } = \frac{ 3x^2 + 4 }{ (x^2 + 1)^2 }\]

This implies that:

Ax^3 + Ax + Bx^2 + B + Cx + D = 3x^2 + 4

By grouping all like terms of the polynomial to compare with the original numerator:

Ax^3 = 0x^3 ---> A = 0
Bx^2 = 3x^2 ---> B = 3
Ax + Cx = 0x --> A = -C = 0
B + D = 4 ------> D = 1

so plugging this back in:

\[\frac{ Ax + B }{ x^2 + 1 } + \frac{ Cx + D }{ (x^2 + 1)^2 } = \frac{ 3 }{ x^2 + 1 } + \frac{ 1 }{ (x^2 + 1)^2 }\]

This means that:

\[\frac{ 3 }{ x^2 + 1 } + \frac{ 1 }{ (x^2 + 1)^2 }\]

is equivalent to

\[\frac{ 3x^2 + 4 }{ (x^2 + 1)^2 }\]