To split in to partial fractions, the expression 1x2(x+a)21x2(x+a)2=Ax+Bx2+C(x+a)+D(x+a)2One method of finding the values of the constants A,B,C &amp; D is as follows.Ax(x+a)2+B(x+a)2+C(x2)(x+a)+Dx2=1Ax(x2+a2+2ax)+B(x2+2ax+a2)+C(x3+ax2)+Dx2=1A(x3+a2x+2ax2)+B(x2+2ax+a2)+C(x3+ax2)+Dx2=1x3(A+C)+x2(2Aa+B+Ca+D)+x(Aa2+2Ba)+Ba2=1Equating coeffecients of x3,x2,x1,x0A+C=0,2Aa+B+Ca+D=0,Aa2+2Ba=0,Ba2=1B=1a2A=−2a3C=2a3D=1a2Is there a shorter method to find the values of A,B,C &amp; D ?

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To split in to partial fractions, the expression 1x2(x+a)21x2(x+a)2=Ax+Bx2+C(x+a)+D(x+a)2One method of finding the values of the constants A,B,C  &amp;amp;  D is as follows.Ax(x+a)2+B(x+a)2+C(x2)(x+a)+Dx2=1Ax(x2+a2+2ax)+B(x2+2ax+a2)+C(x3+ax2)+Dx2=1A(x3+a2x+2ax2)+B(x2+2ax+a2)+C(x3+ax2)+Dx2=1x3(A+C)+x2(2Aa+B+Ca+D)+x(Aa2+2Ba)+Ba2=1Equating coeffecients of x3,x2,x1,x0A+C=0,2Aa+B+Ca+D=0,Aa2+2Ba=0,Ba2=1B=1a2A=−2a3C=2a3D=1a2Is there a shorter method to find the values of A,B,C &amp;amp; D ?

muthe2ulj · Accepted Answer

Ax(x+a)2+B(x+a)2+C(x2)(x+a)+Dx2=1x=0 yields b=1a2x=−a  yields d=1a2Then you can derivate and set again x=−a and x=0 to get A,C or replace B,D in that equation to getAx(x+a)2+C(x2)(x+a)+2x2a2+2axa2+1=1After canceling the 1&#039;s, you can divide by x(x+a) and getA(x+a)+Cx+2a2=0x=0 yields A=−2a3and x=−a yields C=2a3

tutaonana223a · Accepted Answer

In this equation:Ax(x+a)2+B(x+a)2+C(x2)(x+a)+Dx2=1one little thing you can do is to evaluate at x=−a to get0+0+0+D(−a)2=1 to get D. After that can evaluate at x=0 to get B.

To split in to partial fractions, the expression

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