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NCERT Class XII Mathematics Chapter - - Solutions

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..... Miscellaneous Exercise .....
Integrate the functions in Exercises 229 to 252.
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Question : 250 of 271
Marks: +1, -0
x2+x+1(x+1)2(x+2)\frac{x^2+x+1}{(x+1)^2(x+2)}
Solution:  
Let I=∫x2+x+1(x+1)2(x+2)dxI=\int \frac{x^2+x+1}{(x+1)^2(x+2)}dx
Let x2+x+1(x+1)2(x+2)\frac{x^2+x+1}{(x+1)^2(x+2)} =Ax+1+B(x+1)2+Cx+2= \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+2}
∴x2+x+1=A(x+1)(x+2)+B(x+2)+C(x+1)2\therefore x^2 + x + 1 = A(x + 1)(x + 2) + B(x + 2) + C(x + 1)^2 ...(i)
Put x = – 1 in (i), we get
1 – 1 + 1 = B(–1 + 2) ⇒ B = 1
Put x = –2 in (i), we get, 4−2+1=C(−2+1)24 - 2 + 1 = C(-2 + 1)^2 ⇒C=3\Rightarrow C = 3
Put x = 0 in (i)
∴ 1 = 2A + 2B + C ⇒ 1 = 2A + 2 + 3 ⇒ 2A = –4 ⇒ A = –2
⇒I=∫[−2x+1+1(x+1)2+3x+2]dx\Rightarrow I = \int \left[ -\frac{2}{x+1} + \frac{1}{(x+1)^2} + \frac{3}{x+2} \right]dx
=−2log⁡(x+1)+(x+1)−1−1+3log⁡(x+2)+C= -2\log(x+1) + \frac{(x+1)^{-1}}{-1} + 3\log(x+2) + C
=−2log⁡(x+1)−1x+1+3log⁡(x+2)+C= -2\log(x+1) - \frac{1}{x+1} + 3\log(x+2) + C
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