Question

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Answer 1

Given :   ∫ (x²/(x - 1)(x - 2)) dx

To find : Integrate

Solution:

∫x²/(x - 1)(x - 2) dx

= ∫ (x²/(x - 2)   - x²/(x - 1) ) dx

= ∫ (x²/(x - 2) dx -  ∫ x²/(x - 1)  dx

x²/(x - 2)

let say x - 2  = z    => x = z + 2

=> dx  = dz

=  (z + 2)²/z

=  (z² + 4z + 4)/z

= z  + 4  +  4/z

∫ (x²/(x - 2) dx = ∫ (z  + 4  +  4/z)dz    

=  z²/2  + 4z  + 4ln|z|   + C₁

= (x - 2)²/2  + 4(x - 2)  + 4ln|x - 2|   + C₁

x²/(x - 1)

let say x - 1  = y   => x = y + 1

dx = dy

(y + 1)²/y  =  y + 2  + 1/y

∫ x²/(x - 1)  dx = ∫ (y  + 2  +  1/y)dy  

= y²/2 + 2y  + ln|y|  + C₂

= (x - 1)²/2 + 2(x - 1) + ln|x-1| +  C₂

∫ (x²/(x - 2) dx -  ∫ x²/(x - 1)  dx  = (x - 2)²/2  + 4(x - 2)  + 4ln|x - 2|   + C₁ - ((x - 1)²/2 + 2(x - 1) + ln|x-1| +  C₂)

=> ∫(x²/(x - 1)(x - 2)) dx  =  (x - 2)²/2 - (x - 1)²/2 + 4(x - 2)  - 2(x - 1) + 4ln|x - 2|  - ln|x-1| + C

= ( x² - 4x + 4)/2  - (x² -2x + 1)/2 + 4x - 8 -2x + 2 + 4ln|x - 2|  - ln|x-1| + C

= ( -2x + 3)/2  + 2x - 6  +   ln| (x - 2)⁴/(x - 1) |  + C

= x - 9/2 +   ln| (x - 2)⁴/(x - 1) |  + C

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Answer 2

$\textbf{To find:}$

$\int\,\dfrac{x^2}{(x-1)(x-2)}\,dx$

$\textbf{Solution:}$

$\text{Consider,}$

$\dfrac{x^2}{(x-1)(x-2)}$

$=\dfrac{x^2}{x^2-3x+2}$

$=\dfrac{(x^2-3x+2)+3x-2}{x^2-3x+2}$

$=1+\dfrac{3x-2}{(x-1)(x-2)}$

$\text{Now, we resolve \dfrac{3x-2}{(x-1)(x-2)} into partial fractions}$

$\dfrac{3x-2}{(x-1)(x-2)}=\dfrac{A}{x-1}+\dfrac{B}{x-2}$

$3x-2=A(x-2)+B(x-1)$....(1)

$\text{Put x=1 in (1)}$

$1=-A$

$\implies\bf\,A=-1$

$\text{Put x=2 in (1)}$

$4=B$

$\implies\bf\,B=4$

$\implies\dfrac{x^2}{(x-1)(x-2)}=1+\dfrac{(-1)}{x-1}+\dfrac{4}{x-2}$

$\implies\displaystyle\int\dfrac{x^2}{(x-1)(x-2)}\,dx=\int\,1\,dx+\int\dfrac{(-1)}{x-1}\,dx+\int\dfrac{4}{x-2}\,dx$

$\implies\displaystyle\int\dfrac{x^2}{(x-1)(x-2)}\,dx=\int\,1\,dx-\int\dfrac{1}{x-1}\,dx+4\,\int\dfrac{1}{x-2}\,dx$

$\implies\displaystyle\int\dfrac{x^2}{(x-1)(x-2)}\,dx=x-\log|x-1|+4\log|x-2|+c$

$\implies\boxed{\bf\int\dfrac{x^2}{(x-1)(x-2)}\,dx=x+\log\dfrac{|x-2|^4}{|x-1|}+c}$

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