Is this the correct procedure for Integral Partial Fraction.

Question

$∫ (x^3+x^2+x+3)/((x^2+1)(x^2+3))$

The first step I did is distributed the denominator so that I can find out if I should use synthetic division. Which after doing this I discovered that the denominator's exponent is greater therefore I do not use synthetic division

= $∫ (x^3+x^2+x+3)/(x^4+4x^2+3)$ Therfore I solve.

$∫ (x^3+x^2+x+3)/((x^2+1)(x^2+3))$ =∫ $(Ax +B)/(x^2+1) + (Cx+D)/(x^2+3)$ then I distributed $(x^2+1)(x^2+1)$ on both sides and got.

$∫ (x^3+x^2+x+3) $= $(Ax +B)(x^2+3)+( Cx+D)(x^2+1)$

Then I solve for B. x=0

$0+0+0+3 = (A(0)+B)(0+3)+(0+0)(0+1) =$

3=B If this is right how does one find A,CD? Do I make x =0 and B =3 and e verything else 0?

Answer 1

You have to use what is called a coefficient matrix.

So as you said we have that:

$$∫ \frac{x^3+x^2+x+3}{(x^2+1)(x^2+3)}dx =∫ \frac{Ax +B}{x^2+1} + \frac{Cx+D}{x^2+3}dx$$

Then multiplying both sides we get that:

$$x^3+x^2+x+3 = (x^2+3)(Ax+B) + (x^2+1)(Cx+D) = Ax^3+Bx^2+3Ax+3B + Cx^3+Dx^2+Cx+D$$

Now we analyze the coefficients on the LHS and RHS for each power of $x$ and the constants.

So for example we have for $x$ coefficients:

$$3A + C = 1$$

Now, we construct our equations:

For constants:

$$3B+D = 3$$

For $x$ coefficients:

$$3A + C = 1$$

For $x^2$ coefficients:

$$B + D = 1$$

For $x^3$ coefficients:

$$A + C = 1$$

Now construct your matrix and solve for $A$, $B$, $C$, and $D$.

EDIT

You don't have to use a matrix, using simple algebra can yield your answer.

Subtracting the equations gives us $A=D=0$, $B=1$ and $C=1$

Therefore, our partial fraction decomposition is:

$$ \frac{x^3+x^2+x+3}{(x^2+1)(x^2+3)} = \frac{1}{x^2+1} + \frac{x}{x^2+3}$$

Answer 2

We do indeed have $$(Ax +B)(x^2+3)+( Cx+D)(x^2+1) = x^3 + x^2 + x + 3$$

If you put $x=0$, you get $$(A(0) +B)(0 + 3) + (C(0) + D)(0 + 1) = 3 \iff 3B + D = 3$$

Try using $x = (-1)$ and $x = 1$, and apply the same process.

Then use any other constant that hasn't been used yet, say $x = 2$, using the same process. That will guarantee that you have four equations to solve for four unknowns: A, B, C, D.