sjeikdom0

2021-09-26

Find the partial fraction decomposition.

$\frac{{x}^{2}+x-6}{({x}^{2}+1)(x-1)}$

komunidadO

Skilled2021-09-27Added 86 answers

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$\frac{{x}^{2}+x-6}{({x}^{2}+1)(x-1)}$ .

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.

$\frac{{x}^{2}+x-6}{({x}^{2}+1)(x-1)}=\frac{Ax+B}{{x}^{2}+1}+\frac{C}{x-1}$

$=\frac{(Ax+B)(x-1)+C({x}^{2}+1)}{({x}^{2}+1)(x-1)}$

${x}^{2}+x-6=(A+C){x}^{2}+(-A+B)x+(-B+C)$

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:$\frac{{x}^{2}+x-6}{({x}^{2}+1)(x-1)}=\frac{3x+4}{{x}^{2}+1}-\frac{2}{x-1}$

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

Partial fractions for the above function would be of the form

Equating two functions to find the value of A, B and C.

Equating coefficients:

Hence the partial fraction formed is:

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