Partial Fraction Decomposition Problem

Question

I am having trouble with this problem. I need to integrate: $$\frac1{T^4}\times \frac1{K-T}$$ with respect to $T$. If I do PFD:

$$\frac{A}{T^4} + \frac{B}{T^3} + \frac{C}{T^2} +\frac{D}{T} + \frac{E}{K-T}$$

Does this look right?

From here, I am having trouble solving for the coeffs. Can someone help me figure out the method for solving for these coeffs? If I multiply left side denominator across, I just get $1 =$ "a bunch of stuff"

Any help?

Answer 1

Bring the expression to the common denominator $T^4(K-T)$. The numerator is then $$A(K-T)+BT(K-T)+CT^2(K-T)+DT^3(K-T)+ET^4.\tag{1}$$ This is identically equal to $1$. Set $T=0$ in (1). We get $AK=1$, so $A=\frac{1}{K}$. Set $T=K$. We get $EK^4=1$, so $E=\frac{1}{K^4}$.

These were the easiest ones. Now we could expand (1), and find $B,C,D$ by comparing coefficients. Or else we can scan Expression (1). Note that the coefficient of $T$ is $-A+BK$. This is $0$, and we know that $A=\frac{1}{K}$. It follows that $B=\frac{1}{K^2}$. The coefficient of $T^2$ is $-B+CK$. This is $0$, so $C=\frac{1}{K^3}$. One more to go!

Answer 2

\begin{align*} \frac1{T^4(K-T)} &= \frac{A}{T^4} + \frac{B}{T^3} + \frac{C}{T^2} +\frac{D}{T} + \frac{E}{K-T} \\ 1 &= (K-T)A + T(K-T)B + T^2(K-T)C + T^3(K-T)D + T^4E \\ \text{Now equate coefficients} \\ \text{(Unity)} \qquad 1 &= KA \implies A=\tfrac1K \\ (T) \qquad 0 &=-A+KB \implies B = \tfrac1{K^2} \\ (T^2) \qquad 0 &= -B + KC \implies C= \tfrac1{K^3} \\ (T^3) \qquad0 &= -C + KD \implies D = \tfrac1{K^4} \\ (T^4) \qquad0 &= -D + E \implies E = \tfrac1{K^4}\\[12pt] \frac1{T^4(K-T)} &= \frac{1}{KT^4} + \frac{1}{K^2T^3} + \frac{1}{K^3T^2} +\frac{1}{K^4T} + \frac{1}{K^4(K-T)} \\\end{align*}