Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
Question
Integrate the function $\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}$
✓
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) $\Rightarrow \frac{x^{2}+x+1}{(x+1)^{2}(x+2)}$ = $\frac{A(x+1)(x+2)+B(x+2)+C(x+1)^{2}}{(x+1)^{2}(x+2)}$ $\Rightarrow \frac{x^{2}+x+1}{(x+1)^{2}(x+2)}$ = $\frac{A\left(x^{2}+3 x+2\right)+B(x+2)+C\left(x^{2}+2 x+1\right)}{(x+1)^{2}(x+2)}$ ⇒ x2 + x + 1 = Ax2 + 3Ax + 2A + Bx +2B + Cx2 + 2Cx + C ⇒ x2 + x + 1 = (2A + 2B + C) + (3A + B + 2C)x + (A + C)x2 Equating the coefficients of x, x2 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 $\Rightarrow \frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{-2}{x+1}+\frac{1}{(x+1)^{2}}+\frac{3}{(x+2)}$ $\Rightarrow \int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} d x$ = $\int\left(\frac{-2}{x+1}+\frac{1}{(x+1)^{2}}+\frac{3}{(x+2)}\right) d x$ = $-2 . \int\left(\frac{1}{x+1}\right) d x+\int\left(\frac{1}{(x+1)^{2}}\right) d x+3 . \int\left(\frac{1}{(x+2)}\right) d x$ = $-2 \cdot \int\left(\frac{1}{x+1}\right) d x+\int\left((x+1)^{-2}\right) d x+3 \cdot \int\left(\frac{1}{(x+2)}\right) d x$ = $-2 \log |x+1|+\left(\frac{(x+1)^{-1}}{(-1)}\right)+3 \log |x+1|+C$ = $-2 \log |x+1|-\frac{1}{(x+1)}+3 \log |x+1|+C$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.