Question
Evaluate the following integrals:
$\int\frac{\text{x}^2+\text{x}+1}{\text{x}^2+\text{x}+1}\text{dx}$
$\int\frac{\text{x}^2+\text{x}+1}{\text{x}^2+\text{x}+1}\text{dx}$
Comparing the coefficients of like powers of x,
$2=2\lambda$ $\Rightarrow\lambda=1$ $-\lambda+\mu=0$ $\Rightarrow-1+\mu=0$ $\mu=1$ So, $\text{I}_1=\int\frac{(2\text{x}-1)+1}{\text{x}^2-\text{x}+1}\text{ dx}$ $=\int\frac{(2\text{x}-1)}{\text{x}^2-\text{x}+1}\text{ dx}+\int\frac{1}{\text{x}^2-2\text{x}\big(\frac{1}{2}\big)+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2+1}\text{ dx}$ $\text{I}_1=\int\frac{2\text{x}-1}{\text{x}^2-\text{x}+1}\text{ dx+}\int\frac{1}{\big(\text{x}-\frac{1}{2}\big)^2-\Big(\frac{\sqrt3}{2}\Big)^2}\text{ dx}$ $=\log\big|\text{x}^2-\text{x}+1\big|+\frac{2}{\sqrt3}\tan^{-1}\bigg(\frac{\text{x}-\frac{1}{2}}{\frac{\sqrt3}{2}}\bigg)+\text{C}_2$ $\Big[\text{since},\int\frac{1}{\text{x}^2+\text{a}^2}\text{ dx}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$ $=\log\big|\text{x}^2-\text{x}+1\big|+\frac{2}{\sqrt3}\tan^{-1}\Big(\frac{2\text{x}-1}{\sqrt3}\Big)+\text{C}_2\ .....(2)$ Using equation (1) and (2) $\text{I}=\text{x}+\log\big|\text{x}^2-\text{x}+1\big|+\frac{2}{\sqrt3}\tan^{-1}\Big(\frac{2\text{x}-1}{\sqrt3}\Big)+\text{C}$Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.