Integration by parts: $\int{\frac{dx}{(x^2 + a^2)^n}}$.

Question

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.

Answer 1

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*}

Answer 2

Hint:

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

Answer 3

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: $$J = \dfrac1{a^2}\int\dfrac{dx}{(x^2+a^2)^{n-1}} + \dfrac{1}{2a^2(n-1)} \dfrac {x}{(x^2+a^2)^{n-1}} - \dfrac{1}{2a^2(n-1)}\int \dfrac{dx}{(x^2+a^2)^{n-1}}, $$ $$\boxed{J = \dfrac{1}{2a^2(n-1)} \dfrac {x}{(x^2+a^2)^{n-1}} + \dfrac{2n-3}{2a^2(n-1)}\int \dfrac{dx}{(x^2+a^2)^{n-1}}}$$