Integration w/ trig. substitution (x^2 + 1) / (x^2 - 2x + 2)^2

[SpicyPepper]

Another even numbered problem in my book, so no textbook answer. I checked it in WolframAlpha(WA), but the answer came out slightly different. Hopefully no typos in this writeup.

Homework Statement

$$\int \frac{x^2 + 1}{(x^2 - 2x + 2)^2}dx$$

Homework Equations

I factor the denominator:
$$(x^2 - 2x + 2) = (x - 1)^2 + 1$$

Some potential trig. substitution:
$$tan(\Theta) = x - 1$$
$$x = tan(\Theta) + 1$$
$$dx = sec^2(\Theta)d\Theta$$
$$(x - 1)^2 + 1 = sec^2(\Theta)$$

The Attempt at a Solution

$$= \int \frac{x^2 + 1}{((x - 1)^2 + 1)^2}dx$$

$$= \int \frac{(tan^2(\Theta) + 2tan(\Theta) + 1) + 1}{(sec^2(\Theta))^2} sec^2(\Theta)d\Theta$$

$$= \int sin^2(\Theta)d\Theta + 2\int sin(\Theta)cos(\Theta)d\Theta + 2\int cos^2(\Theta)d\Theta$$

$$= \frac{1}{2}\int (1 - cos(2\Theta))d\Theta + 2 \int sin(\Theta)cos(\Theta)d\Theta + 2 \int 1/2(1 + cos(2\Theta))d\Theta$$

$$= \frac{1}{2}\Theta - \frac{1}{4}sin(2\Theta) + sin^2(\Theta) + \Theta + \frac{1}{2}sin(2\Theta) + C$$

$$= \frac{3}{2}arctan(x-1) + \frac{1}{4}2sin(\Theta)cos(\Theta) + sin^2(\Theta) + C$$

$$= \frac{3}{2}arctan(x-1) + (\frac{1}{2})(\frac{x-1}{\sqrt{x^2-2x+2}})(\frac{1}{\sqrt{x^2-2x+2}}) + \frac{(x-1)^2}{x^2-2x+2} + C$$

$$= \frac{3}{2}arctan(x-1) + \frac{2x^2-3x+1}{2(x^2-2x+2)} + C$$

However, plugging it into WA, the answer it spits out is:

$$= -\frac{3}{2}arctan(1-x) + \frac{x-3}{2(x^2-2x+2)} + C$$

I get the 1st term is the same thing, but in the 2nd term, I ended up with 2x^2-3x+1 instead of just x-3. Anyone see a misstep?

I looked through the steps of WA, but I'm not really familiar with what it's saying.

 

[tiny-tim]

Hi SpicyPepper!

$$= \frac{3}{2}arctan(x-1) + \frac{2x^2-3x+1}{2(x^2-2x+2)} + C$$

However, plugging it into WA, the answer it spits out is:

$$= -\frac{3}{2}arctan(1-x) + \frac{x-3}{2(x^2-2x+2)} + C$$

I get the 1st term is the same thing, but in the 2nd term, I ended up with 2x^2-3x+1 instead of just x-3. Anyone see a misstep?

No, the book's C is your C plus one …

always adjust the constant of integration so as to make any fraction a "proper" one.

 

[SpicyPepper]

why is it always so obvious after the fact :p

thx