Question

How you calculate this? `int_0^2(2x^3-6x+9x-5)/(x^2-2x+5)^n`.Suggestion:Forming an odd function by substitution `x-1=t`,like`int_(-1)^1f(t)dt=0`.

Answer 1

See below.

Explanation:

Assuming that the integral is

`int_0^2(2x^3-6x^2+9x-5)/(x^2-2x+5)^n dx` we have that

`2x^3-6x^2+9x-5 = 2(x-1)(x^2-2x+5/2)`

Now, changing variables

`y = x^2-2x+5`

with

`dy = (2x-2)dx`

the integral can be stated as

`int_(y(0))^(y(2)) 2(x-1) (y-5/2)/y^ndy/(2x-2)=int_(y(0))^(y(2))(y-5/2)/y^n dy`

or

`int_(y(0))^(y(2))(y-5/2)/y^n dy=[y^-n ((5 y)/(2 (n-1)) + y^2/(2 - n))]_(y(0))^(y(2))`

Answer 2

Following the suggestion:

Explanation:

Let `t = x-1`, so that `x=t+1` and `dx = dt`.

When `x=0`, `t=-1` and when `x=2`, `t=1`.

The numerator of the integrand becomes

`2x^3-6x^2+9x-5 = 2(t+1)^3-6(t+1)^2+9(t+1)-5`

`= 2(t^3+3t^2+3t+1)-6(t^2+2t+1)+9(t+1)-5`

`= 2t^3+cancel(6t^2)+6t+2-cancel(6t^2)-12t-6+9t+9-5`

`= 2t^3+3t`.

This polynomial is odd.

And the denominator becomes

`(x^2-2x+5)^n = ((t+1)^2-2(t+1)+5)^n`

`= (t^2+2t+1-2t-2+5)^n`

`= (t^2+4)^n`

This polynomial is even, so the quotient if the two polynomials is odd.

Therefore

`int_0^2 (2x^3-6x^2+9x-5)/(x^2-2x+5)^n dx = int_-1^1 (2t^3+3t)/(t^2+4)^n dt`

Because the integrand in the new integral is odd and we are integrating from `-a` to `a`, the integral is `0`.

`int_0^2 (2x^3-6x^2+9x-5)/(x^2-2x+5)^n dx = int_-1^1 (2t^3+3t)/(t^2+4)^n dt = 0`

Bonus note

The substitution also shows us that the graph of the original integrand is symmetric with respect to the point `(1,0)`.

(If `(1+a,b)` is on the graph then `(1-a,-b)` is also on the graph.)