Hi, I have this indefinite integral:
. . . . .\(\displaystyle \displaystyle \int\, \dfrac{-x^2\, +\, 2}{(x^2\, -\, 2x\, +\, 2)^2}\, dx\)
The Ostrogradsky's says that I can solve the integral looking solutions like
\(\displaystyle \displaystyle \int\, \dfrac{P(x)}{Q(x)}\, dx \,=\, \dfrac{P_{1}(x)}{Q_{1}(x)}\, +\, \int \dfrac{P_{2}(x)}{Q_{2}(x)}\, dx\)
where \(\displaystyle \,Q_{1}(x)\,\) is the greatest common divisor of \(\displaystyle \,Q(x)\,\) and \(\displaystyle \,Q'(x)\,\) and \(\displaystyle \,Q_{2}(x)\, =\, \dfrac{Q(x)}{Q_{1}(x)}\)
So \(\displaystyle Q'(x)\, = \,2(2x\, -\, 2)(x^2\,-\,2(x\,-\,1)) \; ; Q_{1}(x)\, =\, x^2\,-\,2x\,+\,2 \; ; Q_{2}(x) \,=\, x^2\,-\,2x\,+\,2\)
The polynomial we are looking for are
\(\displaystyle P_{1}(x)\, =\, -1 \; ; P_{2}(x)\, =\, 0\)
Which steps I need to perform to reach the value of these polynomials?
Thanks in advance
. . . . .\(\displaystyle \displaystyle \int\, \dfrac{-x^2\, +\, 2}{(x^2\, -\, 2x\, +\, 2)^2}\, dx\)
The Ostrogradsky's says that I can solve the integral looking solutions like
\(\displaystyle \displaystyle \int\, \dfrac{P(x)}{Q(x)}\, dx \,=\, \dfrac{P_{1}(x)}{Q_{1}(x)}\, +\, \int \dfrac{P_{2}(x)}{Q_{2}(x)}\, dx\)
where \(\displaystyle \,Q_{1}(x)\,\) is the greatest common divisor of \(\displaystyle \,Q(x)\,\) and \(\displaystyle \,Q'(x)\,\) and \(\displaystyle \,Q_{2}(x)\, =\, \dfrac{Q(x)}{Q_{1}(x)}\)
So \(\displaystyle Q'(x)\, = \,2(2x\, -\, 2)(x^2\,-\,2(x\,-\,1)) \; ; Q_{1}(x)\, =\, x^2\,-\,2x\,+\,2 \; ; Q_{2}(x) \,=\, x^2\,-\,2x\,+\,2\)
The polynomial we are looking for are
\(\displaystyle P_{1}(x)\, =\, -1 \; ; P_{2}(x)\, =\, 0\)
Which steps I need to perform to reach the value of these polynomials?
Thanks in advance
Attachments
Last edited by a moderator: