Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following integrals:$\int\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}}\text{ dx}$
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Answer
Let $\text{I}=\int\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}}\text{ dx}$ $=\int\Big[1+\frac{2\text{x}+1}{\text{x}^2-\text{x}}\Big]\text{dx}$ $=\text{x}+\int\frac{2\text{x}+1}{\text{x}^2-\text{x}}\text{ dx}+\text{C}_1\ ....(1)$ $\text{I}_1=\int\frac{2\text{x}+1}{\text{x}^2-\text{x}}\text{ dx}$ Let $2\text{x}+1=\lambda\frac{\text{d}}{\text{dx}}\big(\text{x}^2-\text{x}\big)+\mu$ $=\lambda(2\text{x}-1)+\mu$ $2\text{x}+1=(2\lambda)\text{x}-\lambda+\mu$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) $\text{I}=\text{x}+\log\big|\text{x}^2+\text{x}\big|+2\log\Big|\frac{\text{x}-1}{\text{x}}\Big|+\text{C}$
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