# Convert the integrand into rational function y=(x^3+2x+1)/(x-1)^3

neela | High School Teacher | (Level 3) Valedictorian

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The integrand y = (x^3+2x+3)/(x-1)^3.

The integrand is  an expression to be integrated.

Alredy the given itegrand is in the form of rational function. The only way we can further change the form is into a sum irredicible forms of k+1/(x-1)+

We put x-1 = u, so that x = u+1.

Then the given integrand is

y = {(u+1)^3+2(u+1)+1}/u^3

y = {u^3+3u^2+3u+1+2(u+1)+1}/u^3

y = {u^3+3u^2+5u+4}/u^3

y = 1+3/u+5/u^2+4/u.

y = 1 + 3/(x-1) + 5/(x-1)^2 + 4/(x-1)^3  is required integrand  in the rarional form.

The itegral of the integrand is I = Interal (integrand) dx

I = Int ydx.

u = x-1

du = dx

Therefore if I =  Integral ydx , then I is given by:

I =  Int ydx = Int {1+3/u+5/u^2+4/u^3}dx

I = u+3logu - 5/2u^1 - 4/3u^2 +C, as Int u^n = (1/n+1)u^(n+1).

I = (x-1) +3log(x-1) -5/2(x-1)  - 2/(x-1)^2 + C.

william1941 | College Teacher | (Level 3) Valedictorian

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We have to find the integrand of y = (x^3 + 2x + 1) / (x-1)^3.

Now first write x as t+1

=> y = (x^3 + 2x + 1) / (x-1)^3

=> y = [(t+1)^3 + 2(t+1) + 1] / t^3

=> y = (t^3 + 3t^2 + 3t + 1 + 2t + 2 + 1) / t^3

Int [y dx ] =  Int [ y dt] as dy = dt

=> Int [(t^3 + 3t^2 + 3t + 1 + 2t + 2 + 1) / t^3]

=> Int [(t^3/ t^3 + 3t^2/t^3 + 3t/t^3 + 1/t^3 + 2t/t^3 + 3/ t^3]

=> Int [ 1 + 3/t + 3/t^2 + 1/t^3 + 2/ t^2 + 3/t^3]

=> Int [ 1 + 3/t + 5/t^2 + 4/t^3]

=>  (3t^2 ln t + t^3 -5t -2) / t^2

=> [3(x-1)^2*ln(x-1) + (x-1)^3 – 5(x-1) -2] / (x-1)^2

Therefore the result is

[3(x-1)^2*ln(x-1) + (x-1)^3 – 5(x-1) -2] / (x-1)^2

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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The integrand is y=(x^3+2x+1)/(x-1)^3.

To convert the integrand into a rational function, we'll use substitution technique.

We'll substitute x-1 = t

We'll differentiate both sides:

dx = dt

We'll express x with respect to t:

x = t + 1

We'll write the integrand in t:

(x^3+2x+1)/(x-1)^3 = [(t+1)^3 + 2(t+1) + 1]/t^5

We'll calculate the indefinite integral

Int [(t+1)^3 + 2(t+1) + 1]/t^5

We'll expand the cube and we'll remove the brackets:

Int (t^3 + 3t^2 + 3t + 1 + 2t + 2 + 1)dt/t^5

We'll combine like terms:

Int (t^3 + 3t^2 + 5t + 4)dt/t^5

We'll use the additive property of inetgrals:

Int (t^3 + 3t^2 + 5t + 4)dt/t^5=Int t^3dt/t^5 + 3Intt^2dt/t^5 + 5Int tdt/t^5 + 4 Int dt/t^5

Int (t^3 + 3t^2 + 5t + 4)dt/t^5=Int dt/t^2+3Int dt/t^3+5Int dt/t^4 + 4 Int dt/t^5

Int dt/t^2 = Int t^-2dt = t^(-2+1)/(-2+1)+C

Int t^-2dt = -1/t

3Int dt/t^3=3Int t^-3dt

3Int t^-3dt = -3/2t^2

5Int dt/t^4 = 5Int t^-4dt

5Int t^-4dt = -5/3t^3

4 Int dt/t^5 = -4/4t^4

4 Int dt/t^5 = -1/t^4

Int (t^3 + 3t^2 + 5t + 4)dt/t^5= -1/t-3/2t^2-5/3t^3-1/t^4 + C