Integrals — Class 12 Maths Solution

Let $I = \int {\frac{{(2x - 1)}}{{(x - 1)(x + 2)(x - 3)}}} dx$

Now $\frac{{2x - 1}}{{(x - 1)(x + 2)(x - 3)}} = \frac{A}{{(x - 1)}} + \frac{B}{{(x + 2)}} + \frac{C}{{(x - 3)}}$

$\Rightarrow$ $2x - 1 = A(x + 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x + 2)$

Let's put $x = 3,$ then
$6 - 1 = C(3 - 1)(3 + 2)$

$\Rightarrow$ $5 = 10C \Rightarrow C = \frac{1}{2}$

Let's put $x = 1$, then
$2 - 1 = A(1 + 2)(1 - 3)$

$\Rightarrow$ $1 = - 6A \Rightarrow A = - \frac{1}{6}$

Now, Let's put $x = - 2,$ then
$- 4 - 1 = B( - 2 - 1)( - 2 - 3)$

$\Rightarrow$ $- 5 = 15B \Rightarrow B = - \frac{1}{3}$

therefore,$I = - \frac{1}{6}\int {\frac{1}{{x - 1}}} dx - \frac{1}{3}\int {\frac{1}{{x + 2}}} dx + \frac{1}{2}\int {\frac{1}{{x - 3}}} dx$

$= - \frac{1}{6}\log |(x - 1)| - \frac{1}{3}\log |(x + 2)| + \frac{1}{2}\log |(x - 3)| + C$

$= - \log |(x - 1){|^{1/6}} - \log |(x + 2){|^{1/3}} + \log |(x - 3){|^{1/2}} + C$

$= \log \left| {\frac{{\sqrt {x - 3} }}{{{{(x - 1)}^{1/6}}{{(x + 2)}^{1/3}}}}} \right| + C$