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I need to show that the following holds using integration by parts:

\begin{equation} \int{\frac{dx}{(x^2 + a^2)^n}} = \frac{x}{2a^2(n-1)(x^2 + a^2)^{n-1}} + \frac{2n - 3}{2a^2(n-1)} \int{\frac{dx}{(x^2 + a^2)^{n-1}}} \end{equation}

I really just don’t know where to start. It’s trivial to construct some solution of the form $\int u'v dx = uv - \int uv' dx$ to the integral on the left, but I can’t see how to get at this exact one.

EDIT:

I have tried to solve it by splitting it up,

\begin{equation} \int{\frac{dx}{(x^2 + a^2)^n}} = \int{\Big( \frac{1}{(x^2 + a^2)^{n-1}} \cdot \frac{1}{(x^2 + a^2)} \Big)}dx \end{equation}

but as far as I can tell this results in something rather different from where I am supposed to end up:

\begin{equation} \int{\Big( \frac{1}{(x^2 + a^2)^{n-1}} \cdot \frac{1}{(x^2 + a^2)} \Big)}dx = \frac{1}{a}arctan \Big(\frac{x}{a}\Big) \cdot \frac{1}{(x^2 + a^2)^{n-1}} + \frac{2(n-1)}{a} \int{\frac{x^2 arctan \big(\frac{x}{a}\big)}{(x^2 + a^2)^{n}}}dx \end{equation}

It is quite possible that I have made a very obvious mistake, so apologies in advance.

I have also tried this:

\begin{equation} \int{(x^2 + a^2)^{-n}dx} = \int{\Big(1 \cdot (x^2 + a^2)^{-n}\Big) dx} = x(x^2 + a^2)^{-n} + n\int{\frac{2x^2}{(x^2 + a^2)^{n+1}} dx} \end{equation}

Again, it doesn’t seem to lead me nearer the specific solution I need.

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  • $\begingroup$ Welcome to math.SE! This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. $\endgroup$ – user37238 Jan 29 '16 at 9:27
  • $\begingroup$ The integration by parts states that $\int u(x) v'(x) \, dx = u(x) v(x) - \int v(x) \, u'(x) dx $, what functions $u$ and $v$ have you tried? $\endgroup$ – user37238 Jan 29 '16 at 9:29
  • $\begingroup$ I updated the post to include two of my attempts. $\endgroup$ – hsherl Jan 29 '16 at 9:48
  • $\begingroup$ I would love to see the trivial solution to the integral on the left. $\endgroup$ – Julián Aguirre Jan 29 '16 at 9:54
  • $\begingroup$ Well I just meant to get something that could be counted as a "solution". That is, a solution here would be anything of the form $uv - \int uv' dx$. Both of the ones included are "solutions" in this broad sense, unless I've made some mistake. I changed the post to clarify, btw. Thanks for pointing it out. $\endgroup$ – hsherl Jan 29 '16 at 10:13
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Let $I_{n}=\int \frac{dx}{(x^{2}+a^{2})^{n}}$, then \begin{align*} I_{n} &= \frac{x}{(x^{2}+a^{2})^{n}}- \int x \, d\left[\frac{1}{(x^{2}+a^{2})^{n}} \right] \\ &=\frac{x}{(x^{2}+a^{2})^{n}}+2n\int \frac{x^{2}dx}{(x^{2}+a^{2})^{n+1}} \\ &=\frac{x}{(x^{2}+a^{2})^{n}}+ 2n\int \left[ \frac{1}{(x^{2}+a^{2})^{n}}-\frac{a^{2}}{(x^{2}+a^{2})^{n+1}} \right] dx \\ &=\frac{x}{(x^{2}+a^{2})^{n}}+ 2nI_{n}-2a^{2}nI_{n+1} \\ 2a^{2}nI_{n+1} &=\frac{x}{(x^{2}+a^{2})^{n}}+(2n-1)I_{n} \\ I_{n+1} &= \frac{x}{2a^{2}n(x^{2}+a^{2})^{n}}+\frac{2n-1}{2a^{2}n}I_{n} \\ I_{n} &= \frac{x}{2a^{2}(n-1)(x^{2}+a^{2})^{n-1}}+\frac{2n-3}{2a^{2}(n-1)}I_{n-1} \end{align*}

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  • $\begingroup$ Thanks you so much. This is really helpful. I'm not entirely sure, however, how do you do the last step there? $\endgroup$ – hsherl Jan 29 '16 at 17:21
  • $\begingroup$ Just replace $n$ by $n-1$. $\endgroup$ – Ng Chung Tak Jan 29 '16 at 18:19
  • $\begingroup$ Huh. Why am I allowed to do that? $\endgroup$ – hsherl Jan 29 '16 at 18:25
  • $\begingroup$ Actually, that was a really silly question. It makes total sense to do that. Thank you, again! This was really helpful. $\endgroup$ – hsherl Jan 29 '16 at 18:38
  • $\begingroup$ For recurrence relation, e.g. $F_{n+2}=F_{n+1}+F_{n}$ and $F_{n}=F_{n-1}+F_{n-2}$ refer to the same sequence if $F_{0}, F_{1}$ are specified. Of course, in the 2nd last $n \neq 0$ and for the last one $n \neq 1$. You may start the derivation from $I_{n-1}$ instead, you won't need to shift $n$ for giving the exact form of your identity posted. Usually, we can't foresee your final conclusion at the beginning. $\endgroup$ – Ng Chung Tak Jan 29 '16 at 18:44
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Hint:

Use integration by parts with $\;\displaystyle \int\dfrac{\mathrm d\mkern1mu x}{(x^2+a^2)^{n-1}}$, setting $$u=\frac1{(x^2+a^2)^{n-1}},\quad\mathrm d\mkern1mu v=\mathrm d\mkern1mu x.$$

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Standart way of such: $$J = \int\dfrac{dx}{(x^2+a^2)^n} = \dfrac1{a^2}\int\dfrac{(a^2+x^2)-x^2}{(x^2+a^2)^n}dx$$ $$J = \dfrac1{a^2}\int\dfrac{dx}{(x^2+a^2)^{n-1}} - \dfrac1{a^2}\int\dfrac{x^2}{(x^2+a^2)^n}dx$$ $$J = \dfrac1{a^2}\int\dfrac{dx}{(x^2+a^2)^{n-1}} + \dfrac{1}{2a^2(n-1)}\int x\,d\, \dfrac1{(x^2+a^2)^{n-1}}.$$ By parts: $$J = \dfrac1{a^2}\int\dfrac{dx}{(x^2+a^2)^{n-1}} + \dfrac{1}{2a^2(n-1)} \dfrac {x}{(x^2+a^2)^{n-1}} - \dfrac{1}{2a^2(n-1)}\int \dfrac{dx}{(x^2+a^2)^{n-1}}, $$ $$\boxed{J = \dfrac{1}{2a^2(n-1)} \dfrac {x}{(x^2+a^2)^{n-1}} + \dfrac{2n-3}{2a^2(n-1)}\int \dfrac{dx}{(x^2+a^2)^{n-1}}}$$

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  • $\begingroup$ How do you go from the first to the second line? $\endgroup$ – hsherl Jan 29 '16 at 17:56
  • $\begingroup$ @hsherl termwise division, thanks for asking $\endgroup$ – Yuri Negometyanov Jan 29 '16 at 23:07

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