Question
Evaluate the following integrals:
$\int\frac{(\text{x}-1)^2}{\text{x}^2+2\text{x}+2}\text{ dx}$
✓
Answer
Let $\int\frac{(\text{x}-1)^2}{\text{x}^2+2\text{x}+2}\text{ dx}$ $=\int\Big(\frac{\text{x}^2-2\text{x}+1}{\text{x}^2-2\text{x}+2}\Big)\text{dx}$ Here,
Therefore,
$\frac{\text{x}^2-2\text{x}+1}{\text{x}^2+2\text{x}+2}=1-\frac{(4\text{x}+1)}{\text{x}^2+2\text{x}+2}\ ....(1)$ Let $4\text{x}+1=\text{A}\frac{\text{d}}{\text{dx}}\big(\text{x}^2+2\text{x}+2\big)+\text{B}$ $4\text{x}+1=\text{A}(2\text{x}+2)+\text{B}$ $4\text{x}+1=(2\text{A})\text{x}+2\text{A}+\text{B}$ Equating Cofficients of like terms $2\text{A}=4$ $\text{A}=2$ $2\text{A}+\text{B}=1$ $2\times2+\text{B}=1$ $\text{B}=-3$ $\int\Big(\frac{\text{x}^2-2\text{x}+1}{\text{x}^2-2\text{x}+2}\Big)\text{dx}$ $=\int\text{dx}-2\int\frac{(2\text{x}+2)}{\text{x}^2+2\text{x}+2}\text{ dx}+3\int\frac{\text{ dx}}{\text{x}^2+2\text{x}+2}$ $=\int\text{dx}-2\int\frac{(2\text{x}+2)}{\text{x}^2+2\text{x}+2}\text{ dx}+3\int\frac{\text{dx}}{(\text{x}+1)^2+1^2}$ $=\text{x}-2\log\big|\text{x}^2+2\text{x}+2\big|+\frac{3}{1}\tan^{-1}\Big(\frac{\text{x}+1}{1}\Big)+\text{C}$ $=\text{x}-2\log\big|\text{x}^2+2\text{x}+2\big|+3\tan^{-1}(\text{x}+1)+\text{C}$Need a full question paper?
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