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Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals1 Mark
Question
Integrate the function $\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}$
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Answer

Given: $\frac{x^{2}+x+1}{(x+1)^{2}(x+2)} $
Let $I=\frac{x^{2}+x+1}{(x+1)^{2}(x+2)} $
Using partial fraction:
Let $\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{(x+2)} .....(i)$
$\Rightarrow \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} = \frac{A(x+1)(x+2)+B(x+2)+C(x+1)^{2}}{(x+1)^{2}(x+2)} $
$\Rightarrow \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} = \frac{A\left(x^{2}+3 x+2\right)+B(x+2)+C\left(x^{2}+2 x+1\right)}{(x+1)^{2}(x+2)} $
$\Rightarrow x^2 + x + 1 = Ax^2 + 3Ax + 2A + Bx +2B + Cx^2 + 2Cx + C$
$\Rightarrow x^2 + x + 1 = (2A + 2B + C) + (3A + B + 2C)x + (A + C)x^{2 }$
Equating the coefficients of $x, x^2$ and constant value. We get:
$A + C = 1$
$3A + B + 2C = 1$
$2A + 2B + C = 1$
After solving we get:
$A = -2, B = 1$ and $C = 3$
$\Rightarrow \frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{-2}{x+1}+\frac{1}{(x+1)^{2}}+\frac{3}{(x+2)} $
$\Rightarrow \int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} d x = \int\left(\frac{-2}{x+1}+\frac{1}{(x+1)^{2}}+\frac{3}{(x+2)}\right) d x $
$ = -2 . \int\left(\frac{1}{x+1}\right) d x+\int\left(\frac{1}{(x+1)^{2}}\right) d x+3 . \int\left(\frac{1}{(x+2)}\right) d x $
$= -2 \cdot \int\left(\frac{1}{x+1}\right) d x+\int\left((x+1)^{-2}\right) d x+3 \cdot \int\left(\frac{1}{(x+2)}\right) d x $
$= -2 \log |x+1|+\left(\frac{(x+1)^{-1}}{(-1)}\right)+3 \log |x+1|+C $
$= -2 \log |x+1|-\frac{1}{(x+1)}+3 \log |x+1|+C $
 

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