How to calculate the following integral without using wolfram mathematica: $$ \int\frac{dx}{(x^2-4x+5)^2} $$
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Set $x-2=t$ and then follow the description here: http://math.stackexchange.com/a/689932/1242 – Hans Lundmark Apr 08 '14 at 13:24
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You can do this without partial fraction decomposition. Note that when completing the square in the denominator, we get: $$x^2 - 4x + 5 = (x - 2)^2 + 1$$
So you'll want to make use of the substitution $x-2 = \tan \theta \implies dx = \sec^2 \theta$.
Then also note that $\tan^2 \theta + 1 = \sec^2\theta$.
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Partial Fraction Decomposition is not required, right if we use Trigonometric Substitution? – lab bhattacharjee Apr 08 '14 at 12:57
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If the function was $\frac{1}{x^2-4x+5}$ then I can solve it: $\int\frac{d(x-2)}{(x-2)^2+1} = \arctan(x-2)$ – Daniel Rusev Apr 08 '14 at 13:14
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HINT: Split the fraction $$\frac{1}{(x^2 - 4x + 5)^2}$$ into partial fractions.
Also, use hints from these notes.
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When you have a factorable polynomial denominator of the form $p_1(x)p_2(x)...p_n(x)$, where $p_i(x)$ are irreducible factors, you can decompose the rational function as such: $$\frac{1}{p_1(x)p_2(x)...p_n(x)}=\sum_{k=1}^n{\frac{A_k}{p_k(x)}}.$$ Thas is partial fraction decomposition. You should be able to integrate that. – Eleven-Eleven Apr 08 '14 at 12:51
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