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Question:

Compute $\displaystyle \int\frac{x^2+1}{(x^2+2)(x+1)} \, dx$

My Approach:

As per my knowledge this integral can be divided in partial Fraction of form $\dfrac{Ax+B}{x^2+px+q}$ and then do the following as per to integrate it.

Solution:

Taking $\dfrac{x^2+1}{(x^2+2)(x+1)}=\dfrac{Ax^2+Bx+C}{x^2+2}+\dfrac{D}{x+1}$

Rule given: Denominator g(x) contains quadratic tractor (may not be factorisable). To each non-repeated quadratic factor of the form $x^2+px+q$(or $x^2+q$, $q$ not equal to $0$), there Should be a partial Fraction of the form $Ax+B/(x^2+px+q)$.

My problem:

I can't understand why the solution provided deviates from the rule that I have studied to solve these kind of problems.

Book:

ISC MATHEMATICS XII

Publishers:

Kalyani

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    $\begingroup$ You should see this to improve your mathematics formatting. $\endgroup$ – Jaideep Khare Jan 26 '18 at 15:54
  • $\begingroup$ Where can I learn about the formatting?And can you edit it for me now? $\endgroup$ – user517784 Jan 26 '18 at 15:57
  • $\begingroup$ Follow the hyperlink on "this" in my previous comment. $\endgroup$ – Jaideep Khare Jan 26 '18 at 15:57
  • $\begingroup$ You haven't told us the rule you have studied, so we can't really explain why. $\endgroup$ – Thomas Andrews Jan 26 '18 at 15:58
  • $\begingroup$ Thanks for notifying. $\endgroup$ – user517784 Jan 26 '18 at 15:58
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Dividing $x^2+1$ by $x^2+2$ yields $1$ as the quotient and $-1$ as the remainder, so we have $$ \frac{x^2+1}{x^2+2} = 1 - \frac 1 {x^2+2}. $$ So \begin{align} & \frac{x^2+1}{(x^2+2)(x+1)} = \frac 1 {x+1} - \frac 1 {(x^2+2)(x+1)} \\[15pt] = {} & \frac 1 {x+1} + \frac{Ax+B}{x^2+2} + \frac{\text{some constant}}{x+1} \\[15pt] = {} & \frac{Ax+B}{x^2+2} + \frac C {x+1} \end{align} Thus I would decompose this into partial fractions in a way consistent with your approach.

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  • $\begingroup$ Your explanation is the best. $\endgroup$ – user517784 Jan 27 '18 at 1:35
  • $\begingroup$ @CalculusProgrammer : I'm glad it helped. $\endgroup$ – Michael Hardy Jan 27 '18 at 3:25
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you must write $$\frac{x^2+1}{(x^2+2)(x+1)}=\frac{Ax+B}{x^2+2}+\frac{C}{x+1}$$

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  • $\begingroup$ Yes absolutely correct.Then is the solution given wrong ? $\endgroup$ – user517784 Jan 26 '18 at 16:00
  • $\begingroup$ your solution is wrong, use my ansatz $\endgroup$ – Dr. Sonnhard Graubner Jan 26 '18 at 16:13
  • $\begingroup$ Okay.Thanks for rectifying it. $\endgroup$ – user517784 Jan 26 '18 at 16:14
  • $\begingroup$ no Problem, it's nice that i could help you $\endgroup$ – Dr. Sonnhard Graubner Jan 26 '18 at 16:15
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From what I understand, the solution in your book points that the fraction can be writeen as

$$ \frac{x^{2}+1}{(x^{2}+2)(x+1)} = \frac{Ax^{2}+Bx+C}{x^{2}+2} + \frac{D}{x+1} $$ $$ = \frac{(Ax^{2}+Bx+C)(x+1) + D(x^{2}+2)}{(x^{2}+2)(x+1)} $$

Of course it can, you just need to find the appropriate constants :

$$ Ax^{3} + (A+B+ D)x^{2} + (B+C)x + (C+2D) = x^{2}+1 $$ $$ \implies A = 0, \:\: A+B+D = 1, \:\: B+C = 0, \:\: C+2D=1$$ Now, actually, from this step, we can already see that $A=0$is a necessity, so the partial fractions must be of the form $$\frac{Bx+C}{x^{2}+2} + \frac{D}{x+1} $$ which is the rule that you already know.

Both the solution and the rule are correct, its just that using the rule will be more efficient.

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