Question
Evaluate the following integrals:$\int\frac{\text{x}^2}{\text{x}^2+7\text{x}+10}\text{ dx}$
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Answer
Let $\text{I}=\int\frac{\text{x}^2}{\text{x}^2+7\text{x}+10}\text{ dx}$ $=\int\Big\{1-\frac{7\text{x}+10}{\text{x}^2+7\text{x}+10}\Big\}\text{dx}$ $=\text{x}-\int\frac{7\text{x}+10}{\text{x}^2+7\text{x}+10}\text{ dx}+\text{C}_1\ ....(1)$ $\text{I}_1=\int\frac{7\text{x}+10}{\text{x}^2+7\text{x}+10}\text{ dx}$ Let $7\text{x}+10=\lambda\frac{\text{d}}{\text{dx}}\big(\text{x}^2+7\text{x}+10\big)+\mu$ $=\lambda(2\text{x}+7)+\mu$ $7\text{x}+10=(2\lambda)\text{x}+7\lambda+\mu$Comparing the coefficients of like powers of x,
$7=2\lambda\Rightarrow\lambda=\frac{7}{2}$ $7\lambda+\mu=10\Rightarrow7\Big(\frac{7}{2}\Big)+\mu=10$ $\mu=-\frac{29}{2}$ So, $\text{I}_1=\int\frac{\frac{7}{2}(2\text{x}+7)-\frac{29}{2}}{\text{x}^2+7\text{x}+10}\text{ dx}$ $\text{I}_1=\frac{7}{2}\int\frac{(2\text{x}+7)}{\text{x}^2+7\text{x}+10}\text{ dx}-\frac{29}{2}\int\frac{1}{\text{x}^2-2\text{x}\big(\frac{7}{2}\big)+\big(\frac{7}{2}\big)^2-\big(\frac{7}{2}\big)^2+10}\text{ dx}$ $\text{I}_1=\frac{7}{2}\int\frac{2\text{x}+7}{\text{x}^2+7\text{x}+10}\text{ dx}-\frac{29}{2}\int\frac{1}{\big(\text{x}+\frac{7}{2}\big)^2-\big(\frac{3}{2}\big)^2}\text{ dx}$ $\text{I}_1=\frac{2}{7}\log\big|\text{x}^2+7\text{x}+10\big|-\frac{29}{2}\times\frac{1}{2\big(\frac{3}{2}\big)}\log\bigg|\frac{\text{x}+\frac{7}{2}-\frac{3}{2}}{\text{x}+\frac{7}{2}+\frac{3}{2}}\bigg|+\text{C}_2$ $\Big[\text{since},\int\frac{1}{\text{x}^2-\text{a}^2}\text{ dx}=\frac{1}{2\text{a}}\log\Big|\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\Big|+\text{C}\Big]$ $\text{I}_1=\frac{7}{2}\log\big|\text{x}^2+7\text{x}+10\big|-\frac{29}{6}\log\Big|\frac{\text{x}+2}{\text{x}+5}\Big|+\text{C}_2\ ....(2)$ Using equation (1) and (2) $\text{I}=\text{x}-\frac{7}{2}\log\big|\text{x}^2+7\text{x}+10\big|+\frac{29}{6}\log\Big|\frac{\text{x}+2}{\text{x}+5}\Big|+\text{C}$
$7=2\lambda\Rightarrow\lambda=\frac{7}{2}$ $7\lambda+\mu=10\Rightarrow7\Big(\frac{7}{2}\Big)+\mu=10$ $\mu=-\frac{29}{2}$ So, $\text{I}_1=\int\frac{\frac{7}{2}(2\text{x}+7)-\frac{29}{2}}{\text{x}^2+7\text{x}+10}\text{ dx}$ $\text{I}_1=\frac{7}{2}\int\frac{(2\text{x}+7)}{\text{x}^2+7\text{x}+10}\text{ dx}-\frac{29}{2}\int\frac{1}{\text{x}^2-2\text{x}\big(\frac{7}{2}\big)+\big(\frac{7}{2}\big)^2-\big(\frac{7}{2}\big)^2+10}\text{ dx}$ $\text{I}_1=\frac{7}{2}\int\frac{2\text{x}+7}{\text{x}^2+7\text{x}+10}\text{ dx}-\frac{29}{2}\int\frac{1}{\big(\text{x}+\frac{7}{2}\big)^2-\big(\frac{3}{2}\big)^2}\text{ dx}$ $\text{I}_1=\frac{2}{7}\log\big|\text{x}^2+7\text{x}+10\big|-\frac{29}{2}\times\frac{1}{2\big(\frac{3}{2}\big)}\log\bigg|\frac{\text{x}+\frac{7}{2}-\frac{3}{2}}{\text{x}+\frac{7}{2}+\frac{3}{2}}\bigg|+\text{C}_2$ $\Big[\text{since},\int\frac{1}{\text{x}^2-\text{a}^2}\text{ dx}=\frac{1}{2\text{a}}\log\Big|\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\Big|+\text{C}\Big]$ $\text{I}_1=\frac{7}{2}\log\big|\text{x}^2+7\text{x}+10\big|-\frac{29}{6}\log\Big|\frac{\text{x}+2}{\text{x}+5}\Big|+\text{C}_2\ ....(2)$ Using equation (1) and (2) $\text{I}=\text{x}-\frac{7}{2}\log\big|\text{x}^2+7\text{x}+10\big|+\frac{29}{6}\log\Big|\frac{\text{x}+2}{\text{x}+5}\Big|+\text{C}$
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