How do you integrate $f (x) = \frac{x - 2}{(x^{2} + 5) (x - 3) (x - 1)}$ $f(x)=(x-2)/((x^2+5)(x-3)(x-1))$ using partial fractions?

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How do you integrate $f (x) = \frac{x - 2}{(x^{2} + 5) (x - 3) (x - 1)}$ $f(x)=(x-2)/((x^2+5)(x-3)(x-1))$ using partial fractions?

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Answer 1 · 2016-12-22T14:00:52

The answer is $= \frac{1}{12} ln (∣ x - 1 ∣) + \frac{1}{28} ln (∣ x - 3 ∣) - \frac{5}{84} ln (x^{2} + 5) - \frac{4}{21 \sqrt{5}} arctan (\frac{x}{\sqrt{5}}) + C$ $=1/12ln(∣x-1∣)+1/28ln(∣x-3∣)-5/84ln(x^2+5)- 4/(21sqrt5)arctan(x/sqrt5)+C$

Explanation:

Let's perform the decomposition into partial fractions

$\frac{x - 2}{(x - 1) (x - 3) (x^{2} + 5)} = \frac{A}{x - 1} + \frac{B}{x - 3} + \frac{C x + D}{x^{2} + 5}$ $(x-2)/((x-1)(x-3)(x^2+5))=A/(x-1)+B/(x-3)+(Cx+D)/(x^2+5)$

$= \frac{A (x - 3) (x^{2} + 5) + B (x - 1) (x^{2} + 5) + (C x + D) (x - 1) (x - 3)}{(x - 1) (x - 3) (x^{2} + 5)}$ $=(A(x-3)(x^2+5)+B(x-1)(x^2+5)+(Cx+D)(x-1)(x-3))/((x-1)(x-3)(x^2+5))$

Therefore,

$x - 2 = A (x - 3) (x^{2} + 5) + B (x - 1) (x^{2} + 5) + (C x + D) (x - 1) (x - 3)$ $x-2=A(x-3)(x^2+5)+B(x-1)(x^2+5)+(Cx+D)(x-1)(x-3)$

Let $x = 1$ $x=1$ , $\Rightarrow$ $=>$ , $- 1 = - 12 A$ $-1=-12A$ , $\Rightarrow$ $=>$ , $A = \frac{1}{12}$ $A=1/12$

Let $x = 3$ $x=3$ , $\Rightarrow$ $=>$ , $1 = 28 B$ $1=28B$ , $\Rightarrow$ $=>$ , $B = \frac{1}{28}$ $B=1/28$

Coefficients of $x^{3}$ $x^3$ , $\Rightarrow$ $=>$ , $0 = A + B + C$ $0=A+B+C$

$C = - \frac{5}{42}$ $C=-5/42$

Let $x = 0$ $x=0$ , $\Rightarrow$ $=>$ , $- 2 = - 15 A - 5 B + 3 D$ $-2=-15A-5B+3D$

$3 D = 15 A + 5 B - 2 = \frac{15}{12} + \frac{5}{28} - 2 = - \frac{4}{7}$ $3D=15A+5B-2=15/12+5/28-2=-4/7$

$D = - \frac{4}{21}$ $D=-4/21$

Therefore,

$\frac{x - 2}{(x - 1) (x - 3) (x^{2} + 5)} = \frac{\frac{1}{12}}{x - 1} + \frac{\frac{1}{28}}{x - 3} + \frac{- \frac{5}{42} x - \frac{4}{21}}{x^{2} + 5}$ $(x-2)/((x-1)(x-3)(x^2+5))=(1/12)/(x-1)+(1/28)/(x-3)+(-5/42x-4/21)/(x^2+5)$

$= \frac{\frac{1}{12}}{x - 1} + \frac{\frac{1}{28}}{x - 3} - \frac{1}{42} \frac{5 x + 8}{x^{2} + 5}$ $=(1/12)/(x-1)+(1/28)/(x-3)-1/42(5x+8)/(x^2+5)$

So,

$I = \int \frac{(x - 2) d x}{(x - 1) (x - 3) (x^{2} + 5)} = \frac{1}{12} \int \frac{d x}{x - 1} + \frac{1}{28} \int \frac{d x}{x - 3} - \frac{1}{42} \int \frac{(5 x + 8) d x}{x^{2} + 5}$ $I=int((x-2)dx)/((x-1)(x-3)(x^2+5))=1/12int(dx)/(x-1)+1/28int(dx)/(x-3)-1/42int((5x+8)dx)/(x^2+5)$

$= \frac{1}{12} ln (∣ x - 1 ∣) + \frac{1}{28} ln (∣ x - 3 ∣) - \frac{1}{42} (5 \int \frac{x d x}{x^{2} + 5} + 8 \int \frac{d x}{x^{2} + 5})$ $=1/12ln(∣x-1∣)+1/28ln(∣x-3∣)-1/42(5int(xdx)/(x^2+5)+8int(dx)/(x^2+5))$

Let $u = x^{2} + 5$ $u=x^2+5$ , $\Rightarrow$ $=>$ , $d u = 2 x d x$ $du=2xdx$

$\int \frac{x d x}{x^{2} + 5} = \frac{1}{2} \int \frac{d u}{u} = \frac{1}{2} ln u = \frac{1}{2} ln (∣ x^{2} + 5 ∣)$ $int(xdx)/(x^2+5)=1/2 int (du)/u=1/2lnu=1/2ln(∣x^2+5∣)$

Let $v = \frac{x}{\sqrt{5}}$ $v=x/sqrt5$ , $\Rightarrow$ $=>$ , $d v = \frac{d x}{\sqrt{5}}$ $dv=dx/sqrt5$

$8 \int \frac{d x}{x^{2} + 5} = \frac{8}{\sqrt{5}} \int \frac{d v}{v^{2} + 1}$ $8int(dx)/(x^2+5)=8/sqrt5int(dv)/(v^2+1)$

Let $v = tan θ$ $v=tantheta$ , $d v = {sec}^{2} θ d θ$ $dv=sec^2theta d theta$

and $1 + {tan}^{2} θ = {sec}^{2} θ$ $1+tan^2theta=sec^2theta$

So,

$\frac{8}{\sqrt{5}} \int \frac{d v}{v^{2} + 1} = \frac{8}{\sqrt{5}} \int (\frac{{sec}^{2} θ}{{sec}^{2} θ} d θ)$ $8/sqrt5int(dv)/(v^2+1)=8/sqrt5int(sec^2theta/sec^2thetad theta)$

$= \frac{8}{\sqrt{5}} θ = \frac{8}{\sqrt{5}} arctan v = \frac{8}{\sqrt{5}} arctan (\frac{x}{\sqrt{5}})$ $=8/sqrt5theta=8/sqrt5arctanv=8/sqrt5arctan(x/sqrt5)$

Putting it all together

$I = \frac{1}{12} ln (∣ x - 1 ∣) + \frac{1}{28} ln (∣ x - 3 ∣) - \frac{5}{84} ln (x^{2} + 5) - \frac{4}{21 \sqrt{5}} arctan (\frac{x}{\sqrt{5}}) + C$ $I=1/12ln(∣x-1∣)+1/28ln(∣x-3∣)-5/84ln(x^2+5)- 4/(21sqrt5)arctan(x/sqrt5)+C$

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How do you integrate $f (x) = \frac{x - 2}{(x^{2} + 5) (x - 3) (x - 1)}$ $f(x)=(x-2)/((x^2+5)(x-3)(x-1))$ using partial fractions?

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Explanation:

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