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