Question

Integrate
with respect to x
x/(x - 1)(x − 2)(x – 3)​

Answer 1

Given Integrand,

$\displaystyle \int \sf \dfrac{x}{(x - 1)(x - 2)(x - 3)} dx$

We know that,

$\boxed{ \boxed{ \displaystyle \int \sf \dfrac{ {ax}^{2} + bx + c}{(x - p)(x - q)(x - r)} dx = \int \bigg( \dfrac{A}{x - p} + \dfrac{B}{x - q} + \dfrac{C}{x - r} \bigg)dx }}$

Therefore,

$\implies \displaystyle \sf \dfrac{x}{(x - 1)(x - 2)(x - 3)} = \dfrac{a}{x - 1} + \dfrac{b}{x - 2} + \dfrac{c}{x - 3} \\ \\ \implies \sf \: \dfrac{x}{ \cancel{(x - 1)(x - 2)(x - 3)}} = \dfrac{a(x - 2)(x - 3) + b(x - 1)(x - 3) + c(x - 2)(x - 1)}{ \cancel{(x - 1)(x - 2)(x -3)}} \\ \\ \implies \sf \: x = a(x - 2)(x - 3) + b(x - 1)(x - 3) + c(x - 2)(x - 1) - - - - (1)$

Putting x = 1 in equation (1),

$\longrightarrow \sf \: 1 = a(1 - 2)(1 - 3) + b(1 - 1)(1 - 3) + c(1 - 2)(1- 1) \\ \\ \longrightarrow \sf \: 1 = 2a \\ \\ \longrightarrow \sf \: a = \dfrac{1}{2}$

Putting x = 2,

$\longrightarrow \sf \: 2= a(2 - 2)(2 - 3) + b(2 - 1)(2 - 3) + c(2 - 2)(2- 1) \\ \\ \longrightarrow \sf \: 2 = - b \\ \\ \longrightarrow \sf \: b = - 2$

Putting x = 3,

$\longrightarrow \sf \: 3= a(3- 2)(3 - 3) + b(3 - 1)(3 - 3) + c(3 - 2)(3- 1) \\ \\ \longrightarrow \sf \: 3 = 2c \\ \\ \longrightarrow \sf \:c = \dfrac{3}{2}$

The integrand can be rewritten as :

$\implies \displaystyle \int \sf \dfrac{x}{(x - 1)(x - 2)(x - 3)}dx = \int \bigg( \dfrac{1}{2(x - 1)} - \dfrac{2}{x - 2} + \dfrac{3}{2(x - 3)} \bigg)dx$

We know that,

$\star \: \boxed{ \displaystyle \int \sf \dfrac{dx}{x} = ln(x) + c}$

Thus,

$\implies \boxed{ \boxed{\sf \: \dfrac{1}{2} ln(x - 1) - 2 ln(x - 2) + \dfrac{3}{2} ln(x - 3) + c}}$

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\sf{\int \dfrac{x}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}dx}$

♣ ᴀɴꜱᴡᴇʀ :

$\boxed{\sf{\int \dfrac{x}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}dx=\dfrac{1}{2}\ln \left|x-1\right|+\dfrac{3}{2}\ln \left|x-3\right|-2\ln \left|x-2\right|+C}}$

♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\bigstar \text { Apply u - substitution: } u=x-2$

Integral Substitution definition :

$\sf{\int \:f\left(g\left(x\right)\right)\cdot \:g^'\left(x\right)dx=\int \:f\left(u\right)du,\:\quad \:u=g\left(x\right)}$

$\mathrm{Substitute:}\:u=x-2$

$\sf{\dfrac{du}{dx} = 1}$

      $\Rightarrow \:du=1dx$

      $\Rightarrow \:dx=1du$

$=\int \dfrac{x}{\left(x-1\right)u\left(x-3\right)}\cdot \:1du$

$\sf{=\int \ffrac{x}{u\left(x-1\right)\left(x-3\right)}du}$

$\bf{u=x-2 \Rightarrow x=u+2}$

$=\int \dfrac{u+2}{u\left(u+2-1\right)\left(u+2-3\right)}du$

$\sf{=\int \dfrac{u+2}{u\left(u+1\right)\left(u-1\right)}du}$

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$\bigstar\text { Expand } \dfrac{u+2}{u(u+1)(u-1)}$

$\text { Apply the fraction rule: } \dfrac{a \pm b}{c}=\dfrac{a}{c} \pm \dfrac{b}{c}$

$\sf{=\dfrac{u}{u\left(u+1\right)\left(u-1\right)}+\dfrac{2}{u\left(u+1\right)\left(u-1\right)}}$

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$\bigstar \mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$\sf{=\int \dfrac{u}{u\left(u+1\right)\left(u-1\right)}du+\int \dfrac{2}{u\left(u+1\right)\left(u-1\right)}du}$

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$\bigstar\: \sf{Simplify\:\int \dfrac{u}{u\left(u+1\right)\left(u-1\right)}du}$

$\mathrm{Cancel\:the\:common\:factor:}\:u$

$=\int \dfrac{1}{\left(u+1\right)\left(u-1\right)}du$

$\text { Expand }(u+1)(u-1): u^{2}-1$

$=\int \dfrac{1}{u^2-1}du$

$u^{2}-1=-\left(-u^{2}+1\right)$

$=\int \dfrac{1}{-\left(-u^2+1\right)}du$

$\text { Take the constant out: } \int a \cdot f(x) d x=a \cdot \int f(x) d x$

$=-\int \dfrac{1}{-u^2+1}du$

$\text { Use the common integral: } \int \dfrac{1}{-y^{2}+1} d u=\dfrac{\ln |u+1|}{2}-\dfrac{\ln |u-1|}{2}$

$\sf{=-\left(\dfrac{\ln \left|u+1\right|}{2}-\dfrac{\ln \left|u-1\right|}{2}\right)}$

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$\bigstar\:\sf{Simplify\:-\left(\dfrac{\ln \left|u+1\right|}{2}-\dfrac{\ln \left|u-1\right|}{2}\right)}$

Distribute parentheses

$=-\left(\dfrac{\ln |u+1|}{2}\right)-\left(-\dfrac{\ln |u-1|}{2}\right)$

Apply minus - plus rules

$-(-a)=a,-(a)=-a$

$=-\dfrac{\ln |u+1|}{2}+\dfrac{\ln |u-1|}{2}$

$=-\dfrac{1}{2} \ln |u+1|+\dfrac{1}{2} \ln |u-1|$

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$\bigstar\:\:\sf{Simplify\:\int \dfrac{2}{u\left(u+1\right)\left(u-1\right)}du}$

$\text{Take the constant out}: \int a \cdot f(x) d x=a \cdot \int f(x) d x =2 \cdot \int \dfrac{1}{u(u+1)(u-1)} d u$

$=2\cdot \int \dfrac{1}{u\left(u+1\right)\left(u-1\right)}du$

$\text { Take the partial fraction of } \dfrac{1}{u(u+1)(u-1)}: \quad-\dfrac{1}{u}+\dfrac{1}{2(u+1)}+\dfrac{1}{2(u-1)}$

$=2\cdot \int \:-\dfrac{1}{u}+\dfrac{1}{2\left(u+1\right)}+\dfrac{1}{2\left(u-1\right)}du$

$\text { Apply the Sum Rule: } \int f(x) \pm g(x) d x=\int f(x) d x \pm \int g(x) d x$

$=2\left(-\int \dfrac{1}{u}du+\int \dfrac{1}{2\left(u+1\right)}du+\int \dfrac{1}{2\left(u-1\right)}du\right)$

$\int \dfrac{1}{u} d u=\ln |u|$

$\int \dfrac{1}{2(u+1)} d u=\dfrac{1}{2} \ln |u+1|$

$\int \dfrac{1}{2(u-1)} d u=\dfrac{1}{2} \ln |u-1|$

$\sf{=2\left(-\ln \left|u\right|+\dfrac{1}{2}\ln \left|u+1\right|+\dfrac{1}{2}\ln \left|u-1\right|\right)}$

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$\bigstar\:\:\bf{2\left(-\ln \left|u\right|+\dfrac{1}{2}\ln \left|u+1\right|+\dfrac{1}{2}\ln \left|u-1\right|\right=-2\ln \left|u\right|+\ln \left|u+1\right|+\ln \left|u-1\right|)}$

$\sf{=-\dfrac{1}{2}\ln \left|u+1\right|+\dfrac{1}{2}\ln \left|u-1\right|-2\ln \left|u\right|+\ln \left|u+1\right|+\ln \left|u-1\right|}$

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$\text { Substitute back } u=x-2$

$=-\dfrac{1}{2}\ln \left|x-2+1\right|+\dfrac{1}{2}\ln \left|x-2-1\right|-2\ln \left|x-2\right|+\ln \left|x-2+1\right|+\ln \left|x-2-1\right|$

$=\dfrac{1}{2}\ln \left|x-1\right|+\dfrac{3}{2}\ln \left|x-3\right|-2\ln \left|x-2\right|$

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$\mathrm{Add\:a\:constant\:to\:the\:solution}$

$\sf{=\dfrac{1}{2}\ln \left|x-1\right|+\dfrac{3}{2}\ln \left|x-3\right|-2\ln \left|x-2\right|+C}$

Why ?

$\sf{if} \:\:\dfrac{d F(x)}{d x}=f(x) \text { then } \int f(x) d x=F(x)+C}$

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