Theory of Intelligence
11TH•Mathematics•Partial Fractions•Question 5

Solution

Resolve into partial fractions: $\dfrac{x^4}{x^4+2x^2+1}$

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Solution

Note:

x4+2x2+1=(x2+1)2x^4+2x^2+1=(x^2+1)^2

So:

x4x4+2x2+1=x4(x2+1)2\frac{x^4}{x^4+2x^2+1}=\frac{x^4}{(x^2+1)^2}

Divide first:

x4(x2+1)2=1+−2x2−1(x2+1)2\frac{x^4}{(x^2+1)^2}=1+\frac{-2x^2-1}{(x^2+1)^2}

Now write:

−2x2−1(x2+1)2=Ax+Bx2+1+Cx+D(x2+1)2\frac{-2x^2-1}{(x^2+1)^2}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}

Solving gives A=0A=0, B=−2B=-2, C=0C=0, D=1D=1.

Hence:

x4x4+2x2+1=1−2x2+1+1(x2+1)2\boxed{\frac{x^4}{x^4+2x^2+1}=1-\frac{2}{x^2+1}+\frac{1}{(x^2+1)^2}}

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