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$$\int \frac1{(x-2)(x^2+1)^2}\,dx= \frac{A}{(x-2)}+ \frac{Bx+C}{(x^2+1)}+ \frac{Dx+E}{(x^2+1)^2}$$

$$1=A(x^2+1)^2 +(Bx+C)(x-2)(x^2+1)+(Dx+E)(x-2)$$

Am I doing this correct , I'm getting too many equations and I cant solve them?? (and yes its not my homework but exercising for exam).

Michael Hardy
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Endrit Shabani
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    The coefficient of $A$ is wrong. – User1234 Jun 03 '18 at 14:16
  • It should be $A(x^2+1)^2$. – User1234 Jun 03 '18 at 14:17
  • yes i forgot to add it here in the paper I wrote it correct – Endrit Shabani Jun 03 '18 at 14:17
  • Rather than saying $$ \int \frac1{(x-2)(x^2+1)^2},dx= \frac{A}{(x-2)}+ \frac{Bx+C}{(x^2+1)}+ \frac{Dx+E}{(x^2+1)^2},$$ you should say either $$ \int \frac1{(x-2)(x^2+1)^2},dx= \int \frac{A}{(x-2)}+ \frac{Bx+C}{(x^2+1)}+ \frac{Dx+E}{(x^2+1)^2} ,dx$$ or $$ \frac1{(x-2)(x^2+1)^2} = \frac{A}{(x-2)}+ \frac{Bx+C}{(x^2+1)}+ \frac{Dx+E}{(x^2+1)^2}. $$ The two expressions you've connected with an "equals" sign are certainly not equal. – Michael Hardy Jun 03 '18 at 14:33

4 Answers4

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I'll teach you a "new" easy way of solving those $$\frac{1}{(x-2)(x^2+1)^2}=\frac{A}{x-2}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{(x^2+1)^2}$$

$$1=A(x^2+1)^2+(Bx+C)(x-2)(x^2+1)+(Dx+E)(x-2)$$ Now substitute $x=2$, you get $A=\frac{1}{25}$

Substitute $x=i$, you get $1=(Di+E)(i-2)=-D-2E+Ei-2Di \implies E=2D \,\land D+2E=-1 $ $\implies D=\frac{-1}{5} \,\land E=\frac{-2}{5} $

Substitute $x=1$, you get $1=4A-2B-2C-D-E \implies \frac{-3}{25}=B+C$

Substitute $x=0$, you get $1=A-2C-2E \implies C=\frac{-2}{25}$

So we have our solution $A=\frac{1}{25}$, $B=-\frac{1}{25}$, $C=-\frac{2}{25}$, $D=-\frac{1}{5}$, $E=-\frac{2}{5}$.

Jakobian
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  • On a side note: Also, when the roots of the polynomial in the denominator are of multiplicity 1, you can, using derivatives, derive a formula for such decompositions. – Jakobian Jun 03 '18 at 14:35
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It should be $$1=A(x^2+1)^2+(Bx+C)(x-2)(x^2+1)+(Dx+E)(x-2)$$

Dr. Sonnhard Graubner
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After correcting ($A(x^2+1)\rightarrow A(x^2+1)^2$), you get the system$$\left\{\begin{array}{l}A+B=0\\-2B+C=0\\-2 A + B - 2 C + D=0\\-2 B + C - 2 D +E=0\\A - 2 C - 2 E=1.\end{array}\right.$$It has five equations and five unknows, besides being linear. What's the problem?

José Carlos Santos
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Looks right. Why too many equations?

The sum of the $x^4$ terms will add to $0$, sum of the $x^3$ terms will add to $0$, the sum of the $x^2$ terms will add to $0$, the sum of the $x$ terms will add to zero, and the sum of the constants will add to $1$. Five equations, five unknowns.

Greg
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