Question

Evaluate : 'int(3x-2)/(x^(2)-3x +2)dx'​

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\displaystyle\int\sf \dfrac{3x - 2}{( {x}^{2} - 3x + 2)} \: dx$

can be rewritten as

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

So to evaluate this integral, we first resolve by partial fraction

$\rm :\longmapsto\:Let \: \dfrac{3x - 2}{(x - 1)(x - 2)} = \dfrac{a}{x - 1} + \dfrac{b}{x - 2} - - - (1)$

$\rm :\longmapsto\:3x - 2 = a(x - 2) + b(x - 1) - - (2)$

On substituting x = 1 in equation (2), we get

$\rm :\longmapsto\:3(1) - 2 = a(1 - 2) + b(1 - 1)$

$\rm :\longmapsto\:3 - 2 = a( - 1)$

$\bf\implies \:a = - \: 1 - - (3)$

On substituting x = 2, in equation (2), we get

$\rm :\longmapsto\:3(2) - 2 = a(2- 2) + b(2- 1)$

$\rm :\longmapsto\:6 - 2 = b(1)$

$\bf\implies \:b = 4 - - - (4)$

Now,

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

$\rm :\longmapsto\: \dfrac{3x - 2}{(x - 1)(x - 2)} = \dfrac{ - 1}{x - 1} + \dfrac{4}{x - 2}$

On integrating both sides w. r. t. x, we get

$\rm :\longmapsto\: \displaystyle\int\sf \dfrac{3x - 2}{(x - 1)(x - 2)}dx =\displaystyle\int\sf \dfrac{ - 1}{x - 1}dx + \displaystyle\int\sf \dfrac{4}{x - 2}dx$

$\rm :\longmapsto\: \displaystyle\int\sf \dfrac{3x - 2}{(x - 1)(x - 2)}dx = - log(x - 1) + log(x - 2) + c$

$\: \: \: \: \: \: \: \: \: \: \: \green{\boxed{ \because \: \bf \: \displaystyle\int\sf\dfrac{1}{x} \: dx = log(x) + c}}$

$\bf :\longmapsto\: \displaystyle\int\bf \dfrac{3x - 2}{(x - 1)(x - 2)}dx = log \bigg | \frac{x - 2}{x - 1} \bigg| + c$

$\: \: \: \: \: \: \: \: \: \green{\boxed{ \because \bf \: logx - logy = log \frac{x}{y} }}$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Factor \: in \: denominator & \bf Decomposition \: in \: partial \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf ax + b & \sf \displaystyle \frac{A}{{ax + b}} \\ \\ \sf {(ax + b)}^{2} & \sf \displaystyle \frac{{{A_1}}}{{ax + b}} + \frac{{{A_2}}}{{{{\left( {ax + b} \right)}^2}}} \\ \\ \sf {ax}^{2} + bx + c & \sf \displaystyle \frac{{Ax + B}}{{a{x^2} + bx + c}} \end{array}} \\ \end{gathered}$