how to solve this into partial fractions

Question

I'm having a bit of a hard time putting this into partial fractions: $$\frac{10}{x^2+2x+1+\pi^2}.$$ I know that the roots of the denominator are $-1 \pm i\pi$, but I dont know how to proceed on puting this in partial fractions. Can anyone here help me, please?

Answer 1

You get:

$\frac{10}{x^2-2x+1+\pi^2}=\frac{A}{x+(1-\pi i)}+\frac{B}{x+(1+\pi i)}=\frac{(A+B)x+A(1+\pi i)+B(1-\pi i)}{x^2-2x+1+\pi^2}$

So you have to solve the following system: $\left\{\begin{matrix} A+B=0\\ A+B+(A-B)\pi i=10 \end{matrix}\right.$

Which is equivalent to: $\left\{\begin{matrix} B=-A\\ (A-B)\pi i=10 \end{matrix}\right.$

The solution is: $\left\{\begin{matrix} 2A\pi i=10 \Rightarrow A=\frac{5}{\pi i}=-\frac{5 i}{\pi}\\ B=-A \Rightarrow B=\frac{5 i}{\pi} \end{matrix}\right.$

So you end up with: $\frac{10}{x^2-2x+1+\pi^2}=-\frac{\frac{5 i}{\pi}}{x+(1-\pi i)}+\frac{\frac{5 i}{\pi}}{x+(1+\pi i)}$

Answer 2

\begin{gather*} I=\frac{10}{x^{2} +2x+1+\pi ^{2}} =\frac{10}{( x+1+i\pi )( x+1-i\pi )}\\ =\frac{10}{2i\pi }\frac{2i\pi }{( x+1+i\pi )( x+1-i\pi )}\\ =\frac{10}{2i\pi }\frac{( x+1+i\pi ) -( x+1-i\pi )}{( x+1+i\pi )( x+1-i\pi )}\\ =\frac{10}{2i\pi }\left(\frac{1}{( x+1-i\pi )} -\frac{1}{( x+1+i\pi )}\right) \end{gather*}

Answer 3

Let the fraction be $$\frac{A}{x+1-i\pi}+\frac{B}{x+1+i\pi}.$$ Then $$A(x+1+i\pi) + B(x+1-i\pi) \equiv 10.$$ Compare the coefficient of $x$ on the LHS to see that $A = -B$. Then $2Ai\pi = 10$, so that $A = \dfrac{5}{i\pi}.$ Hence $$\frac{10}{x^2+2x+1+\pi^2} \equiv \frac{5}{i\pi}\left(\frac{1}{x+1-i\pi}-\frac{1}{x+1+i\pi}\right).$$

P.S. My answer simplified the fraction.

Answer 4

Welcome to MSE:$$\frac{10}{x^2+2x+1+\pi^2}=\frac{10}{(x+1)^2+\pi^2}=\\ \frac{a}{x+1+i\pi}+\frac{b}{x+1-i\pi}=\\\frac{a(x+1-i\pi)+b(x+1+i\pi)}{(x+1)^1+\pi^2}=\\ \frac{(x+1)(a+b)-i\pi(a-b)}{(x+1)^1+\pi^2}=\\\to a+b=0 \to a=-b\\ and \\(-i\pi)(a-b)=10\to a-b=\frac{10}{-i\pi}$$so $$-2b=\frac{10}{-i\pi}\to b=\frac{5}{i\pi}=\frac{5}{i\pi}\frac ii=-\frac{5i}{\pi}\\a=\frac{5i}{\pi}$$

Answer 5

Remember than you can always process by identification of the coefficients as a fallback when you don't see some other shortcut.

$\dfrac{10}{x^2+2x+1+\pi^2}=\dfrac{a}{x-1+i\pi}+\dfrac{b}{x-1-i\pi}=\dfrac{x(a+b)-(a+b+ai\pi-bi\pi)}{x^2+2x+1+\pi^2}$

Therefore you get to solve

$\begin{cases}a+b=0\\(a+b)+(a-b)i\pi=-10\end{cases}\iff\begin{cases}b=-a\\2ai\pi=-10\end{cases}\iff\begin{cases}a=\frac {5i}{\pi}\\b=-\frac{5i}{\pi}\end{cases}$