- #1
SpicyPepper
- 20
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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.
\int \frac{x^2 + 1}{(x^2 - 2x + 2)^2}dx
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)
= \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.
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.
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