Verified answer

[tex]\large\underline{\sf{Solution-}}[/tex]

Given integral is

[tex]\rm \: \displaystyle{\bf{\;\;{\int \frac{ {x}^{2}dx}{(x - 1)(x - 2)} }}} \\ [/tex]

To evaluate this integral, we use method of Partial Fractions.

So, Let assume that

[tex]\rm \: \dfrac{ {x}^{2} }{(x - 1)(x - 2)} = 1 + \dfrac{a}{x - 1} + \dfrac{b}{x - 2} - - - (1) \\ [/tex]

On taking LCM, we get

[tex]\rm \: {x}^{2} = (x - 1)(x - 2) + a(x - 2) + b(x - 1) \\ [/tex]

Substituting x = 1, in above expression, we get

[tex]\rm \: 1 = a(1 - 2) \\ [/tex]

[tex]\rm \: - a = 1 \\ [/tex]

[tex]\rm\implies \: \boxed{ \rm{ \:\: a \: = \: - \: 1 \: \: }}\\ [/tex]

On substituting x = 2, in the above expression, we get

[tex]\rm \: 4 = b(2 - 1) \\ [/tex]

[tex]\rm\implies \: \boxed{ \rm{ \:\: b \: = \: 4 \: \: }}\\ [/tex]

On substituting the values of a and b in equation (1), we get

[tex]\rm \: \dfrac{ {x}^{2} }{(x - 1)(x - 2)} = 1 - \dfrac{1}{x - 1} + \dfrac{4}{x - 2} \\ [/tex]

On integrating both sides, we get

[tex]\rm \: \displaystyle\int\rm \dfrac{ {x}^{2} }{(x - 1)(x - 2)}dx = \displaystyle\int\rm 1dx - \displaystyle\int\rm \dfrac{1}{x - 1}dx + \displaystyle\int\rm \dfrac{4}{x - 2}dx \\ [/tex]

[tex]\rm \: \displaystyle\int\rm \dfrac{ {x}^{2} }{(x - 1)(x - 2)}dx =x - log |x - 1| + 4log |x - 2| + c \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information :-

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{\displaystyle\int\sf \frac{dx}{ {x}^{2} + {a}^{2} } = \dfrac{1}{a} {tan}^{ - 1} \dfrac{x}{a} + c }\\ \\ \bigstar \: \bf{\displaystyle\int\sf \frac{dx}{ \sqrt{ {x}^{2} - {a}^{2} } } = log |x + \sqrt{ {x}^{2} - {a}^{2} } | + c }\\ \\ \bigstar \: \bf{\displaystyle\int\sf \frac{dx}{ \sqrt{ {a}^{2} - {x}^{2} } } = {sin}^{ - 1} \frac{x}{a} + c }\\ \\ \bigstar \: \bf{\displaystyle\int\sf \frac{dx}{ \sqrt{ {x}^{2} + {a}^{2} } } = log |x + \sqrt{ {x}^{2} + {a}^{2}} | + c}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]