Indefinite Integrals — Maths STD 12 Science — Question

$\text{I}=\int\frac{(\text{x}^2+\text{x}+1)}{(\text{x}+1)^2(\text{x}+2)}\ \text{dx}$

Let, $\int\frac{\text{x}^2+\text{x}+1}{(\text{x}+1)^2(\text{x}+2)}=\frac{\text{A}}{\text{x}+1}+\frac{\text{B}}{(\text{x}+1)^2}+\frac{\text{C}}{\text{x}+2}$

$\Rightarrow\frac{\text{x}^2+\text{x}+1}{(\text{x}+1)^2(\text{x}+2)}=\frac{\text{A}(\text{x}+1)(\text{x}+2)+\text{B}(\text{x}+2)+\text{C}(\text{x}+1)^2}{(\text{x}+1)^2(\text{x}+2)}$

$\Rightarrow\text{x}^2+\text{x}+1=\text{A}(\text{x}^2+\text{x}+2\text{x}+2)\\+\text{Bx}+2\text{B}+\text{C}(\text{x}^2+2\text{x}+1)$

$\Rightarrow\text{x}^2+\text{x}+1=(\text{A}+\text{C})\text{x}^2+(3\text{A}+\text{B}+2\text{C})\text{x}\\+(2\text{A}+2\text{B}+\text{C})$

Equating coefficient of like terms,

$\text{A}+\text{C}=1\ ...(1)$

$3\text{A}+\text{B}+2\text{C}=1\ ...(2)$

$2\text{A}+2\text{B}+\text{C}=1\ ...(3)$

Solving these three equation we get

$\text{A}=-2$

$\text{B}=1$

$\text{C}=3$

Hence, $\frac{\text{x}^2+\text{x}+1}{(\text{x}+1)^2(\text{x}+2)}=\frac{-2}{\text{x}+1}+\frac{1}{(\text{x}+1)^2}+\frac{3}{\text{x}+2}$

$\therefore\text{I}=-2\int\frac{\text{dx}}{\text{x}+1}+\int\frac{\text{d}}{(\text{x}+1)^2}+3\int\frac{\text{dx}}{\text{x}+2}$

$=-2\log|\text{x}+1|-\frac{1}{\text{x}+1}+3\log|\text{x}+2|+\text{C}$