Solve $\int \frac{dx}{(ax^2+bx+c)^k}$. Is it easy? For example solve $\int \frac {dx}{(x^2+1)^2}$.
I mean the antiderivatve from the integral.
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http://en.wikipedia.org/wiki/Integration_by_reduction_formulae – lab bhattacharjee Aug 29 '14 at 13:00
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1Hint: For the first complete the square and make a substitution: $$ax^2+bx+c=a\left(x-\frac{b}{2a}\right)^2-\left(\frac{b^2}{4a}+c\right)\equiv ax'^2+c'$$ Kind of result: http://www.wolframalpha.com/input/?i=Integrate%5B1%2F%28x%C2%B2-a%29%5Ek%2Cx%5D – Matthias Aug 29 '14 at 13:08
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Yes, It is easy. sorry and thanks. – user161341 Aug 29 '14 at 13:12
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This is doable using elementary techniques for all integers and half-integers $k$, though in practice this can be involved by the standards of a first-term calculus course. Try, for example $(x^2 + 1)^{3/2}$. – Travis Willse Aug 29 '14 at 13:42
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Hint: Let $$I(c)=\int\frac{dx}{\underbrace{ax^2+bx+c}_{P(x)}}=\frac2{\sqrt{-\Delta}}\arctan\frac{P'(x)}{\sqrt{-\Delta}},$$ where $\Delta=b^2-4ac$. If $\Delta>0$, use the fact that $~\arctan(iu)=i\cdot\text{arctanh }u=\dfrac i2\cdot\ln\dfrac{1+u}{1-u}$ . Now, by differentiating under the integral sign, we have $$I'(c)=-\int\frac{dx}{(ax^2+bx+c)^2},\quad I''(c)=2\int\frac{dx}{(ax^2+bx+c)^3},$$ and in general, $$I^{(n)}(c)=(-1)^n\cdot n!\cdot\int\frac{dx}{(ax^2+bx+c)^{n+1}}=\dfrac{\eth^n}{\eth c^n}\left[\frac2{\sqrt{-\Delta}}\arctan\frac{P'(x)}{\sqrt{-\Delta}}\right].$$
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Lucian, please recall, and again review the Enforcement of Quality Standards – amWhy Jan 19 '22 at 16:51