# Finding a General Pattern for the Partial fraction of $\frac{1}{{{\left( 1+x \right)}^{n}}\left( 1+{{x}^{2}} \right)}$

I am interested in finding the general pattern for the partial fraction of: $$\frac{1}{{{\left( 1+x \right)}^{n}}\left( 1+{{x}^{2}} \right)}$$ where $$n=1,2,3,......$$ here is the partial fractions from $$n=1\quad to\quad 10$$ $$\left( \begin{matrix} 1 & \frac{1-u}{2\left( {{u}^{2}}+1 \right)}+\frac{1}{2(u+1)} \\ 2 & -\frac{u}{2\left( {{u}^{2}}+1 \right)}+\frac{1}{2(u+1)}+\frac{1}{2{{(u+1)}^{2}}} \\ 3 & \frac{-u-1}{4\left( {{u}^{2}}+1 \right)}+\frac{1}{4(u+1)}+\frac{1}{2{{(u+1)}^{2}}}+\frac{1}{2{{(u+1)}^{3}}} \\ 4 & \frac{1}{2{{(u+1)}^{3}}}+\frac{1}{2{{(u+1)}^{4}}}-\frac{1}{4\left( {{u}^{2}}+1 \right)}+\frac{1}{4{{(u+1)}^{2}}} \\ 5 & \frac{u-1}{8\left( {{u}^{2}}+1 \right)}-\frac{1}{8(u+1)}+\frac{1}{4{{(u+1)}^{3}}}+\frac{1}{2{{(u+1)}^{4}}}+\frac{1}{2{{(u+1)}^{5}}} \\ 6 & \frac{u}{8\left( {{u}^{2}}+1 \right)}-\frac{1}{8(u+1)}-\frac{1}{8{{(u+1)}^{2}}}+\frac{1}{4{{(u+1)}^{4}}}+\frac{1}{2{{(u+1)}^{5}}}+\frac{1}{2{{(u+1)}^{6}}} \\ 7 & \frac{u+1}{16\left( {{u}^{2}}+1 \right)}-\frac{1}{16(u+1)}-\frac{1}{8{{(u+1)}^{2}}}-\frac{1}{8{{(u+1)}^{3}}}+\frac{1}{4{{(u+1)}^{5}}}+\frac{1}{2{{(u+1)}^{6}}}+\frac{1}{2{{(u+1)}^{7}}} \\ 8 & -\frac{1}{8{{(u+1)}^{3}}}-\frac{1}{8{{(u+1)}^{4}}}+\frac{1}{4{{(u+1)}^{6}}}+\frac{1}{2{{(u+1)}^{7}}}+\frac{1}{2{{(u+1)}^{8}}}+\frac{1}{16\left( {{u}^{2}}+1 \right)}-\frac{1}{16{{(u+1)}^{2}}} \\ 9 & \frac{1-u}{32\left( {{u}^{2}}+1 \right)}+\frac{1}{32(u+1)}-\frac{1}{16{{(u+1)}^{3}}}-\frac{1}{8{{(u+1)}^{4}}}-\frac{1}{8{{(u+1)}^{5}}}+\frac{1}{4{{(u+1)}^{7}}}+\frac{1}{2{{(u+1)}^{8}}}+\frac{1}{2{{(u+1)}^{9}}} \\ 10 & -\frac{u}{32\left( {{u}^{2}}+1 \right)}+\frac{1}{32(u+1)}+\frac{1}{32{{(u+1)}^{2}}}-\frac{1}{16{{(u+1)}^{4}}}-\frac{1}{8{{(u+1)}^{5}}}-\frac{1}{8{{(u+1)}^{6}}}+\frac{1}{4{{(u+1)}^{8}}}+\frac{1}{2{{(u+1)}^{9}}}+\frac{1}{2{{(u+1)}^{10}}} \\ \end{matrix} \right)$$ Can any body see the pattern if any, Can we write it as a sum??

Problem background: I am trying to find a closed form for the integral: $$\int{\frac{1}{{{\left( 1+x \right)}^{n}}\left( 1+{{x}^{2}} \right)}dx}$$

• Perhaps I'm quite late, but I just saw this. With the substitution $x=\frac{1-t}{1+t}$ you will get: $$\int{\frac{1}{{{\left( 1+x \right)}^{n}}\left( 1+{{x}^{2}} \right)}dx}=-\frac{1}{2^n} \int \frac{(1+t)^n}{1+t^2}dt$$ Now use binomial expansion :) But before I would suggest to write $t^2+1=(t+i)(t-i)$. Commented Jul 26, 2019 at 22:52
• it is never too late...
– logo
Commented Jul 27, 2019 at 2:56

Lemma. For $$|x+1| < \sqrt{2}$$, we have $$\begin{gathered} \sum_{k=0}^{\infty} \cos(k\pi/4)\left(\frac{x+1}{\sqrt{2}}\right)^k = \frac{1-x}{1+x^2}, \\ \sum_{k=0}^{\infty} \sin(k\pi/4)\left(\frac{x+1}{\sqrt{2}}\right)^k = \frac{1+x}{1+x^2}. \end{gathered}$$

In particular, using $$\cos(x) + \sin(x) = \sqrt{2}\sin(x+\frac{\pi}{4})$$, we obtain

$$\sum_{k=0}^{\infty} \frac{\sin((k+1)\pi/4)}{2^{(k+1)/2}} (x+1)^k = \frac{1}{1+x^2}$$

for $$|x+1| < \sqrt{2}$$. Plugging this to OP's rational function,

$$\frac{1}{(1+x)^n(1+x^2)} = \left( \sum_{k=1}^{n} \frac{\sin(k\pi/4)}{2^{k/2}} \frac{1}{(x+1)^{n+1-k}} \right) + \sum_{k=0}^{\infty} \frac{\sin((n+k+1)\pi/4)}{2^{(n+k+1)/2}} (x+1)^k.$$

The latter sum can be further simplified by using the addition formula for $$\sin$$, yielding

\begin{align*} \sum_{k=0}^{\infty} \frac{\sin((n+k+1)\pi/4)}{2^{(n+k+1)/2}} (x+1)^k &= \frac{\sin((n+1)\pi/4)}{2^{(n+1)/2}} \frac{1-x}{1+x^2} + \frac{\cos((n+1)\pi/4)}{2^{(n+1)/2}} \frac{1+x}{1+x^2} \\ &= \frac{\cos(n\pi/4)}{2^{n/2}} \frac{1}{1+x^2} - \frac{\sin(n\pi/4)}{2^{n/2}} \frac{x}{1+x^2} \end{align*}

Combining altogether, we get

$$\frac{1}{(1+x)^n(1+x^2)} = \left( \sum_{k=1}^{n} \frac{\sin(k\pi/4)}{2^{k/2}} \frac{1}{(x+1)^{n+1-k}} \right) + \frac{\cos(n\pi/4)-x \sin(n\pi/4)}{2^{n/2}(1+x^2)}.$$

Although this is shown initially on the region $$|x+1| < \sqrt{2}$$, this continues to hold everywhere since any two rational functions which coincide at infinitely many points must be equal.

• Ok this is perfect!!!!! where you found this lemma???
– logo
Commented Jul 15, 2019 at 16:33
• @logo it is the general series expansion centred around the point x = -1, which can be done using complex partial fractions, or, if you want to have additional difficulties, you can inductively establish the general term and remain inside the real field. Commented Jul 16, 2019 at 3:22
• Now its clear..thank you
– logo
Commented Jul 16, 2019 at 11:11

Put $$\begin{equation*} f(x)=\dfrac{1}{(1+x)^n(1+x^2)}. \end{equation*}$$ Then the partial fraction of $$f(x)$$ has the form $$\begin{equation*} f(x) =\dfrac{ax+b}{1+x^2} + \sum_{k=1}^{n}\dfrac{c_k}{(1+x)^{n+1-k}}\tag {1} \end{equation*}$$ or $$\begin{equation*} f(x) = \dfrac{d_1}{x+i}+\dfrac{d_2}{x-i}+\sum_{k=1}^{n}\dfrac{c_k}{(1+x)^{n+1-k}} \end{equation*}$$ where $$\begin{gather*} d_1= \underset{x=-i}{\rm{res}}f(x) = \dfrac{1}{(1-i)^n(-2i)}=-\dfrac{e^{i\frac{n\pi}{4}}}{\sqrt{2}^{n}2i}\\[2ex] d_2= \underset{x=i}{\rm{res}}f(x) = \dfrac{1}{(1+i)^n(2i)}=\dfrac{e^{-i\frac{n\pi}{4}}}{\sqrt{2}^{n}2i}. \end{gather*}$$ However, $$\begin{gather*} a=d_1+d_2 =-\dfrac{\sin\left(\frac{n\pi}{4}\right)}{2^{\frac{n}{2}}}\\[2ex] b=i(d_2-d_1)= \dfrac{\cos\left(\frac{n\pi}{4}\right)}{2^{\frac{n}{2}}}. \end{gather*}$$ Furthermore $$\begin{gather*} c_k = \underset{x=-1}{\rm{res}}(1+x)^{n-k}f(x)= \underset{x=-1}{\rm{res}}\dfrac{1}{(1+x)^k(1+x^2)}=\\[2ex] \left.\dfrac{1}{(k-1)!}\dfrac{d^{k-1}}{dx^{k-1}}\left(\dfrac{1}{2i}\left(\dfrac{1}{x-i}-\dfrac{1}{x+i}\right)\right)\right|_{x=-1} =\\[2ex] \dfrac{1}{2i(k-1)!}(-1)^{k-1}(k-1)!\left(\dfrac{1}{(-1-i)^{k}}-\dfrac{1}{(-1+i)^{k}}\right)= \dfrac{1}{2i}\left(\dfrac{1}{(1-i)^{k}}-\dfrac{1}{(1+i)^{k}}\right)=\\[2ex] \dfrac{1}{2i}\left(\dfrac{e^{i\frac{k\pi}{4}}}{\sqrt{2}^{k}}-\dfrac{e^{-i\frac{k\pi}{4}}}{\sqrt{2}^{k}}\right) = \dfrac{\sin\left(\frac{k\pi}{4}\right)}{2^{\frac{k}{2}}} \end{gather*}$$ Now we know all coefficients in (1).

• the power of residues ....thanks and +1
– logo
Commented Jul 15, 2019 at 21:51

I'm going to carry this almost all the way through and stop just before the final step because I got tired of all the necessary details.

$$\begin{array}\\ f_n(x) &=\frac{1}{{{\left( 1+x \right)}^{n}}\left( 1+{{x}^{2}} \right)}\\ &=\frac{a+bx}{1+x^2}+\sum_{k=1}^n \frac{c_k}{(1+x)^k}\\ g_n(x) &=f_n(x)(1+x^2)(1+x)^n\\ &=1\\ &=(a+bx)(1+x)^n+\sum_{k=1}^n c_k(1+x^2)(1+x)^{n-k}\\ g_n(-1) &=2c_n\\ c_n &=\frac12\\ g_n(i) &=1\\ &=(a+bi)(1+i)^n\\ &=2^{n/2}(a+bi)(\frac{1+i}{\sqrt{2}})^n\\ &=2^{n/2}(a+bi)(e^{i\pi/4})^n\\ &=2^{n/2}(a+bi)e^{ni\pi/4}\\ a+bi &=2^{-n/2}e^{-ni\pi/4}\\ &=2^{-n/2}(\cos(-n\pi/4)+i\sin(-n\pi/4))\\ &=2^{-n/2}(1, \frac{1-i}{\sqrt{2}}, -i, -\frac{1+i}{\sqrt{2}}, -1, -\frac{1-i}{\sqrt{2}}, i, \frac{1+i}{\sqrt{2}}) \quad\text{for }n\equiv (0,1,2,3,4,5,6,7)\bmod 8 \\ &=(\frac1{2^{n/2}}, \frac{1-i}{2^{(n+1)/2}}, -\frac{i}{2^{n/2}}, -\frac{1+i}{2^{(n+1)/2}}, -\frac1{2^{n/2}}, -\frac{1-i}{2^{(n+1)/2}}, \frac{i}{2^{n/2}}, \frac{1+i}{2^{(n+1)/2}})\\ &=(\frac1{2^{4m}}, \frac{1-i}{2^{4m+1}}, -\frac{i}{2^{4m+1}}, -\frac{1+i}{2^{4m+2}}, -\frac1{2^{4m+2}}, -\frac{1-i}{2^{4m+3}}, \frac{i}{2^{4m+3}}, \frac{1+i}{2^{4m+4}}) \quad n=8m+k, k=0...7\\ &=(\frac1{2^{4m}}, \frac{1-i}{2^{4m+1}}, \frac{-i}{2^{4m+1}}, \frac{-1-i}{2^{4m+2}}, \frac{-1}{2^{4m+2}}, \frac{-1+i}{2^{4m+3}}, \frac{i}{2^{4m+3}}, \frac{1+i}{2^{4m+4}})\\ &=\dfrac1{2^{\lceil n/2 \rceil}}(1, 1-i, -i, -1-i, -1, -1+i, i, 1+i)\\ g_n^{(j)}(x) &=0 \qquad\text{for } j \ge 1\\ &=((a+bx)(1+x)^n)^{(j)}+\sum_{k=1}^n c_k((1+x^2)(1+x)^{n-k})^{(j)}\\ &=u^{(j)}(x)+\sum_{k=1}^n c_kv_k^{(j)}(x)\\ u^{(j)}(x) &=((a+bx)(1+x)^n)^{(j)}\\ &=\sum_{h=0}^j \binom{j}{h}(a+bx)^{(h)}(1+x)^n)^{(j-h)}\\ (a+bx)^{(h)} &=a+bx, b, 0, ... \qquad\text{for }h = 0, 1, 2, ...\\ ((1+x)^n)^{(h)} &=\frac{n!}{(n-h)!}(1+x)^{n-h}\\ v_k^{(j)}(x) &=((1+x^2)(1+x)^{n-k})^{(j)}(x)\\ &=\sum_{h=0}^j \binom{j}{h}(1+x^2)^{(h)}((1+x)^{n-k})^{(j-h)}(x)\\ (1+x^2)^{(h)} &=1+x^2, 2x, 2, 0, ... \qquad \text{for } h=0, 1, 2, ...\\ ((1+x)^{n-k})^{(h)}(x) &=\frac{(n-k)!}{(n-k-h)!}(1+x)^{n-k-h}\\ \end{array}$$

And at this point, I'll stop.