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Question Bank [2022] — Maths (commerce) STD 12 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Question Bank [2022]4 Marks
Question
$\int \frac{x}{(x-1)^2(x+2)} d x$
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Answer

Let $I =\int \frac{x}{(x-1)^2(x+2)} d x$
Let $\frac{x}{(x-1)^2(x+2)}=\frac{ A }{x-1}+\frac{ B }{(x-1)^2}+\frac{ C }{(x+2)}$
$\therefore x = A ( x -1)( x +2)+ B ( x +2)+ C ( x -1)^2 \ldots\text{(i)}$
Putting $x=1$ in (i), we get
$ 1= A (0)(3)+ B (3)+ C (0)^2$
$\therefore 1=3 B$
$\therefore B =\frac{1}{3} $ Putting $x=\square 2$ in (i), we get
$ -2=A(-3)(0)+B(0)+C(9)$
$\therefore-2=9 C $
$\therefore C=-\frac{2}{9}$
Putting $x=-1$ in (i), we get
$ -1= A (-2)(1)+ B (1)+ C (4)$
$\therefore-1=-2 A +\frac{1}{3}-\frac{8}{9}$
$\therefore-1=-2 A -\frac{5}{9}$
$\therefore 2 A =-\frac{5}{9}+1=\frac{4}{9}$
$\therefore A =\frac{2}{9}$
$\therefore \frac{x}{(x-1)^2(x+2)}=\frac{\frac{2}{9}}{x-1}+\frac{\frac{1}{3}}{(x-1)^2}+\frac{\left(-\frac{2}{9}\right)}{x+2}$
$\therefore I =\int\left[\frac{2}{9} \frac{\frac{1}{3}}{(x-1)^2}+\frac{\left(-\frac{2}{9}\right)}{x+2}\right] d x $
$=\frac{2}{9} \int \frac{1}{x-1} d x+\frac{1}{3} \int(x-1)^{-2} d x-\frac{2}{9} \int \frac{1}{x+2} d x$
$=\frac{2}{9} \log |x-1|+\frac{1}{3} \cdot \frac{(x-1)^{-1}}{-1}-\frac{2}{9} \log |x+2|+ c$
$=\frac{2}{9} \log |x-1|-\frac{2}{9} \log |x+2|-\frac{1}{3} \times \frac{1}{(x-1)}+ c$
$\therefore I =\frac{2}{9} \log \left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+ c $

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