Using partial fraction expression, find x(t) for X(s) =...

Transcript

00:01 Let's start the solution in this question.

00:03 We have given integration x square minus 6 upon x cube minus x dx then question says that by using partial fraction solve this integration.

00:15 So this let this integration is equal to i so it can be written as x square minus 6 upon take x common here.

00:24 We get x x square minus 1 integration dx.

00:29 So now it can be written as i equal to integration a upon x plus bx plus c upon x square minus 1 dx.

00:43 So now we try to find the value of a and b and c by using partial fraction so it can be written as x square minus 6 upon x x square minus 1 which is equal to a upon x plus bx plus c upon x square minus 1 now take lcm here.

01:10 So we get a x square minus 1 plus bx plus c into x upon x x square minus 1 equal to x square minus 6 upon x x square minus 1 here.

01:27 This one is cancelled out.

01:28 So we get here x square minus 6 equal to multiply this a in a bracket.

01:35 We get a x square minus a plus multiply this x in a bracket.

01:40 We get b x square plus cx.

01:43 So now here we get x square minus 6 equal to a x square plus b x square plus cx minus a now by comparing here we get coefficient of x square is 1 so we get a plus b equal to 1 and coefficient of x is here 0 so we get cx is equal to 0 not cx only c equal to 0 because here coefficient of x is 0 and here coefficient of x is c so we get c is 0 and here minus a is a constant and here minus 6 is a constant.

02:22 So we get minus a equal to minus 6...

Question

Using partial fraction expression, find x(t) for \(X(s) = \frac{s(s+1)}{(s+2)(s+3)(s+4)}\) [10 Marks]

Question

Using partial fraction expression, find x(t) for \(X(s) = \frac{s(s+1)}{(s+2)(s+3)(s+4)}\) [10 Marks]

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