How do you use partial fraction decomposition to decompose the fraction to integrate $\frac{x^{2} + x - 6}{(x^{2} + 1) (x - 1)}$ $(x^2+x-6)/((x^2+1)(x-1))$ ?

Question

How do you use partial fraction decomposition to decompose the fraction to integrate $\frac{x^{2} + x - 6}{(x^{2} + 1) (x - 1)}$ $(x^2+x-6)/((x^2+1)(x-1))$ ?

1 Answer

Impact of this question

1818 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-22T03:23:03

You have a second-degree factor and a first-degree factor in the denominator. $x^{2} + 1$ $x^2 + 1$ can't be factored any further as all real numbers, so this will separate into:

$\frac{A}{x - 1} + \frac{B x + C}{x^{2} + 1}$ $A/(x - 1) + (Bx+C)/(x^2 + 1)$

Cross-multiply:

$= \frac{A (x^{2} + 1) + (B x + C) (x - 1)}{(x - 1) (x^{2} + 1)}$ $= (A(x^2 + 1) + (Bx+C)(x-1))/((x-1)(x^2+1))$

Multiply it out (and arrange it to look descending if you want):
$= \frac{A x^{2} + A + B x^{2} - B x + C x - C}{(x - 1) (x^{2} + 1)}$ $= (Ax^2 + A + Bx^2 - Bx + Cx - C)/((x-1)(x^2+1))$

However you do it, you should see now that there are $x^{2}$ $x^2$ , $x$ $x$ , and $x^{0}$ $x^0$ (constant) terms.
$= \frac{(A + B) x^{2} + (- B + C) x + (A - C)}{(x - 1) (x^{2} + 1)}$ $= (color(highlight)((A+B))x^2 + color(highlight)((-B + C))x + color(highlight)((A - C)))/((x-1)(x^2+1))$

I purposefully changed $- (B - C)$ $- (B - C)$ to $+ (- B + C)$ $+ (-B + C)$ to match the original numerator's adding $x$ $x$ . We can equate them back to the original equation's numerator terms' coefficients.

1) $A + B = 1$ $A + B = 1$
2) $- B + C = 1$ $-B + C = 1$
3) $A - C = - 6$ $A - C = -6$

Now, here's what I would do. The parentheses indicate which equation was used when. Single parentheses indicate newly-numbered equations.

$A = 1 - B$ $A = 1 - B$ (1)
$1 - B - C = - 6$ $1 - B - C = -6$ (1 $\to$ $->$ 3)
4) $B = 7 - C$ $B = 7 - C$

$C - 7 + C = 1$ $C - 7 + C = 1$ (4 $\to$ $->$ 2)
$2 C = 8$ $2C = 8$
$C = 4$ $color(green)(C = 4)$

$A - 4 = - 6$ $A - 4 = -6$ (3)
$A = - 2$ $color(green)(A = -2)$

$- 2 + B = 1 \to B = 3$ $-2 + B = 1 -> color(green)(B = 3)$ (1)

Plug them back in and integrate:

$\Rightarrow \int - \frac{2}{x - 1} d x + \int \frac{3 x + 4}{x^{2} + 1} d x$ $=> int-2/(x-1)dx + int(3x + 4)/(x^2 + 1)dx$

Notice how you can separate the second integral (some people may have trouble thinking of that):
$= - 2 \int \frac{1}{x - 1} d x + \int \frac{3 x}{x^{2} + 1} d x + 4 \int \frac{1}{x^{2} + 1} d x$ $= -2int1/(x-1)dx + int(3x)/(x^2 + 1)dx + 4int 1/(x^2 + 1)dx$

And prepare the now-second integral for u-substitution by multiplying by $\frac{3}{2} \cdot \frac{2}{3}$ $3/2*2/3$ since $\frac{2}{3} \cdot 3 x = 2 x$ $2/3*3x = 2x$ and $d (x^{2}) = 2 x d x$ $d(x^2) = 2xdx$ :

$= - 2 \int \frac{1}{x - 1} d x + \frac{3}{2} \int \frac{2 x}{x^{2} + 1} d x + 4 \int \frac{1}{x^{2} + 1} d x$ $= -2int1/(x-1)dx + 3/2int(2x)/(x^2 + 1)dx + 4int 1/(x^2 + 1)dx$

For the first integral, it is $- 2 ln | x - 1 |$ $-2ln|x-1|$ . With the second one, letting $u = x^{2} + 1$ $u = x^2+1$ , $d u = 2 x d x$ $du = 2xdx$ (that's what we were preparing for) and thus it becomes:

$\frac{3}{2} \int \frac{1}{u} d u$ $3/2int1/udu$

so it is therefore $\frac{3}{2} ln ∣ ∣ x^{2} + 1 ∣ ∣$ $3/2ln|x^2+1|$ . The third is $4 arctan x$ $4arctanx$ since $4 \frac{d}{d x} [arctan x] = 4 \cdot \frac{1}{1 + x^{2}}$ $4d/(dx)[arctanx] = 4*1/(1+x^2)$ .

$\Rightarrow - 2 ln | x - 1 | + \frac{3}{2} ln ∣ ∣ x^{2} + 1 ∣ ∣ + 4 arctan x + C$ $=> color(blue)(-2ln|x-1| + 3/2ln|x^2+1| + 4arctanx + C)$

Science

Math

Humanities

... and beyond

How do you use partial fraction decomposition to decompose the fraction to integrate $\frac{x^{2} + x - 6}{(x^{2} + 1) (x - 1)}$ $(x^2+x-6)/((x^2+1)(x-1))$ ?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you use partial fraction decomposition to decompose the fraction to integrate (x^2+x-6)/((x^2+1)(x-1))x2+x−6(x2+1)(x−1)(x^2+x-6)/((x^2+1)(x-1))?

1 Answer

Related questions

Impact of this question

How do you use partial fraction decomposition to decompose the fraction to integrate $\frac{x^{2} + x - 6}{(x^{2} + 1) (x - 1)}$ $(x^2+x-6)/((x^2+1)(x-1))$ ?