Find integral from 0 to 2 of:

 

X²

^{_______________}

X² - 2X + 2

Is this homework?

 

If so, what have you done so far to try to solve this?

 

I can give you a hint if you are really stuck.

Edited April 28, 20169 yr by ajb

No, it isn't some assignment - I'm a teacher trying to stay on top of it; but -as everyone knows - I'm also victim of the same conspiracy of the churches oppressing & smearing the rock & rollers for decades, & this makes it so that I don't always have any professor or fellow colleague I can bounce thing off of. I've tried integration by parts, substitution (symbolic), partial fractions integration; I also tried completing the square - & there may be something going in that direction. I already solved another similar problem, finding the integral of

 

X

________________

X² - 2X + 2

 

Any instruction from the "Expert" would be awesome!

Let Y=X-1. You now have an integrand (Y+1)²/(Y²+1)=1+2Y/(Y²+1). (Note dY=dX)

The indefinite integral is Y + ln(Y²+1). Final answer 2(1+ln(2)).

The method I would use in this example us to think about the derivative of the log of your deniminator. You see that this is almost what you are looking for. But you can then make a simple 'correction' to get the indefinite integral you require (trying adding 1 as this correction)

 

 

You can then substitute in the value 0 and 2 to get your definite integral.

 

 

P.S. You can always check the answer using Wolfram Alpha.

Edited April 29, 20169 yr by ajb

I used the following:-

 

[math] \int \frac{x^2}{x^2-2x+2} dx = \int \frac{x^2-2x+2+2x-2}{x^2-2x+2} dx = \int 1+\frac{2x-2}{x^2-2x+2} dx [/math]

 

Then the integration of the fraction is straightforward as the top is the derivative of the bottom.

Edited May 19, 20169 yr by Juno

[latex] \int\limits^2_0 \dfrac{x^2}{x^2-2x+2} dx [/latex]

 

[latex] =\int\limits^2_0 \dfrac{x^2-2x+2+2x-2}{x^2-2x+2} dx [/latex]

 

[latex] =\int\limits^2_0 1+ \dfrac{2x-2}{x^2-2x+2} dx [/latex]

 

[latex] =\int\limits^2_0 dx+\int\limits^2_0 \dfrac{2x-2}{x^2-2x+2} dx [/latex]

 

Let [latex] t=x^2-2x+2 [/latex]

[latex] dt=(2x-2)dx [/latex]

 

[latex] =\bigg[x\bigg]^2_0+\int\limits^2_0\dfrac{dt}{t} [/latex]

 

[latex] =2+\bigg[\ln\,t\bigg]^2_0\;\;\implies\bigg[\ln(x^2-2x+2)\bigg]^2_0 [/latex]

 

[latex] =2+\bigg[\ln\,2-\ln\,2\bigg] [/latex]

[latex] =2+0=2 [/latex]

Edited November 16, 20169 yr by deesuwalka