`(t^6 + 1)/(t^6 + t^3)` Write out the form of the partial fraction decomposition of the function. Do not determine the numerical values of the coefficients.

Textbook Question

Chapter 7, 7.4 - Problem 6a - Calculus: Early Transcendentals (7th Edition, James Stewart).
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kspcr111 | In Training Educator

Posted on

Ohk here is the solution for it

`(t^6 + 1)/(t^6 + t^3) `

sol:

`(t^6 + 1)/(t^6 + t^3)`

we have to divide the `(t^6 + 1)` with `(t^6 + t^3)` by long division process

Then we get

quotient = 1

and remanider = `1-t^3`

so ,

`(t^6 + 1)/(t^6 + t^3)` = `1+ (1-t^3)/(t^6 + t^3)`

`= 1- (t^3 -1)/(t^6 + t^3) `

`= 1- (t-1)(t^2+t+1)/(t^3(t+1)(t^2 -t+1)) `

Now the partial decomposition is done using it as follows:

=> `=1- ((t-1)(t^2+t+1))/(t^3(t+1)(t^2 -t+1))`

=` 1- [(A/t)+(B/t^2) +(C/t^3) +D/(t+1) + (Et+F)/(t^2 -t+1)]`

is the desired partial decomposition

:)

kspcr111 | In Training Educator

Posted on

`(t^6 + 1)/(t^t + t^3)`

sol:

As the denominator is a polynomial of "t" th term we cannot find the partial fraction

we can simplify it as follows

`(t^6 + 1)/(t^t + t^3)`

=`(t^6 )/(t^t + t^3) +( 1)/(t^t + t^3)`

= `1/(t^(t-6)+ (1/(t^3)))+( 1)/(t^t + t^3)`

:)