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2021-09-26

Find the partial fraction decomposition.
$\frac{{x}^{2}+x-6}{\left({x}^{2}+1\right)\left(x-1\right)}$

Step 1
Formations of Partial fractions is method to separate a fraction into its components such that when the components are added together, they give back the original fraction.
This method is very useful in integration or while performing Fourier transforms, Laplace transforms of various functions .
Step 2
The given function is represented as $\frac{{x}^{2}+x-6}{\left({x}^{2}+1\right)\left(x-1\right)}$.
Partial fractions for the above function would be of the form $\frac{Ax+B}{{x}^{2}+1}+\frac{C}{x-1}$
Equating two functions to find the value of A, B and C.
$\frac{{x}^{2}+x-6}{\left({x}^{2}+1\right)\left(x-1\right)}=\frac{Ax+B}{{x}^{2}+1}+\frac{C}{x-1}$
$=\frac{\left(Ax+B\right)\left(x-1\right)+C\left({x}^{2}+1\right)}{\left({x}^{2}+1\right)\left(x-1\right)}$
${x}^{2}+x-6=\left(A+C\right){x}^{2}+\left(-A+B\right)x+\left(-B+C\right)$
Equating coefficients:
$A+C=1$
$-A+B=1$
$-B+C=-6$
$B+C=2$
$2C=-4$
$C=-2$
$A=3$
$B=4$
Hence the partial fraction formed is: $\frac{{x}^{2}+x-6}{\left({x}^{2}+1\right)\left(x-1\right)}=\frac{3x+4}{{x}^{2}+1}-\frac{2}{x-1}$

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