So I have to integrate $$ \frac{3x+2}{(x)(x+1)^3} $$ so I have the terms $$ \frac{A}{x} + \frac{B}{x+1} +\frac{C}{(x+1)^2} +\frac{D}{(x+1)^3} $$ ... and then I have $$ 3x+2=A(x+1)^3 + B(x)(x+1)^2 +C(x)(x+1)^2+ D(x)(x+1)^2 $$ ... here I tried to equal the quotients and I have $A=2$..how do I find the others?
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From $$ \frac{A}{x} + \frac{B}{x+1} +\frac{C}{(x+1)^2} +\frac{D}{(x+1)^3} $$ you have: $$\frac{A(x+1)^3+Bx(x+1)^2+Cx(x+1)+Dx}{x(x+1)^3}=\frac{3x+2}{x(x+1)^3}$$ Now let $x=0$ then $$A(0+1)^3=3\times0+2\to A=2$$ Let $x=-1$ then $$D(-1)=3\times(-1)+2=-1\to D=-1$$ Set $x=1$ then $$8A+4B+2C+D=3\times1+2=5\to 16+4B+2C-1=5\to4B+2C=-10$$ Set $x=2$ then $$27A+18B+6C+2D=3\times 2+2=8\to 54+18B+6C-2=8\to18B+6C=-44$$ Now solve the two final equation respect to $B$ and $C$.
Mikasa
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You’re starting off on the wrong foot. When you put
$$\frac{A}{x} + \frac{B}{x+1} +\frac{C}{(x+1)^2} +\frac{D}{(x+1)^3}\tag{1}$$
back together into a single fraction, that fraction must have the same denominator as the original one. $(1)$ becomes
$$\frac{A(x+1)^3+Bx(x+1)^2+Cx(x+1)+Dx}{x(x+1)^3}\;.$$
and you want to choose $A,B,C$, and $D$ so that
$$\frac{A(x+1)^3+Bx(x+1)^2+Cx(x+1)+Dx}{x(x+1)^3}=\frac{3x+2}{x(x+1)^3}\;.$$
Two fractions with the same denominator are equal if and only their numerators are equal, so you require that
$$A(x+1)^3+Bx(x+1)^2+Cx(x+1)+Dx=3x+2\;.\tag{2}$$
Setting $x=0$, you find that $A=2$. Setting $x=-1$, you find that $-D=-1$, so $D=1$. Thus, $(2)$ reduces to
$$2(x+1)^3+Bx(x+1)^2+Cx(x+1)+x=3x+2\;,$$ or
$$2(x+1)^3+Bx(x+1)^2+Cx(x+1)=2x+2\;.$$ Now just multiply out the lefthand side:
$$2x^3+6x^2+6x+2+Bx^3+2Bx^2+Bx+Cx^2+Cx=2x+2\;,$$ or
$$(B+2)x^3+(2B+C+6)x^2+(B+C+4)x=0\;.$$
The coefficients on the righthand side are all $0$, so all that remains is to solve
$$\left\{\begin{align*} &B+2=0\\ &2B+C+6=0\\ &B+C+4=0 \end{align*}\right.$$
for $B$ and $C$. The first equation makes this easy: $B=-2$, and then $C=-2$ as well.
Brian M. Scott
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Hint $\ \ $ Evaluating the equation $\rm\ A(x\!+\!1)^3\!+Bx(x\!+\!1)^2\!+Cx(x\!+\!1)+Dx\:=\:3x\!+\!2 $ $$\begin{eqnarray} \rm at\ \ x&=&\ \ \ 0 &&\Rightarrow\rm\ \ \ \color{#C00}{A = 2}\\ \rm at\ \ x&=& {-}1 &&\Rightarrow\rm\ \ \ \color{#C00}{D = 1}\\ \rm at\ \ x&=&\ \ \ 1 &&\Rightarrow\rm\ \ \ 4B +\rm 2C &=& -12\\ \rm at\ \ x&=&-2&&\Rightarrow\rm -2B +\rm2C &=& 0\end{eqnarray}$$
Subtracting the last two yields $\rm\ 6B = -12\:\Rightarrow\: \color{#C00}{B = -2}\:\Rightarrow\: \color{#C00}{C = -2}$
Remark $\ $ Generally a polynomial of degree $n$ over $\Bbb Q$ is determined uniquely by its values at $n+1$ points. We judiciously chose the above points in order to keep the arithmetic simple.
The first half is the Heaviside cover-up method, which can also be applied to nonlinear factors.
