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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following integrals:$\int\frac{(1-\text{x}^2)}{\text{x}(1-2\text{x})}\text{ dx}$
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Answer

Let $\text{I}=\int\frac{1-\text{x}^2}{\text{x}(1-2\text{x})}\text{ dx}$ $=\int\frac{1-\text{x}^2}{\text{x}-2\text{x}^2}\text{ dx}$ $=\int\frac{1-\text{x}^2}{2\text{x}^2-\text{x}}\text{ dx}$ $=\int\bigg[\frac{1}{2}+\frac{\frac{\text{x}}{2}-1}{2\text{x}^2-\text{x}}\bigg]\text{dx}$ $\text{I}=\frac{1}{2}\text{x}+\int\frac{\frac{\text{x}}{2}-1}{2\text{x}^2-\text{x}}\text{ dx}+\text{C}_1\ ....(1)$ $\text{I}_1=\int\frac{2\text{x}+1}{\text{x}^2-\text{x}}\text{ dx}$ Let $\frac{\text{x}}{2}-1=\lambda\frac{\text{d}}{\text{dx}}\big(2\text{x}^2-\text{x}\big)+\mu$ $=\lambda(4\text{x}-1)+\mu$ $\frac{\text{x}}{2}-1=(4 \lambda)\text{x}-\lambda+\mu$Comparing the coefficients of like powers of x,
$\frac{1}{2}=4\lambda\Rightarrow\lambda=\frac{1}{8}$ $-\lambda+\mu=-1\Rightarrow-\Big(\frac{1}{8}\Big)+\mu=-1$ $\mu=-\frac{7}{8}$ So, $\text{I}_1=\int\frac{\frac{1}{8}(4\text{x}-1)-\frac{7}{8}}{2\text{x}^2-\text{x}}\text{ dx}$ $\text{I}=\frac{1}{8}\int\frac{4\text{x}-1}{2\text{x}^2-\text{x}}\text{ dx}-\frac{7}{8}\int\frac{1}{2\big(\text{x}^2-\frac{\text{x}}{2}\big)}\text{ dx}$ $\text{I}=\frac{1}{8}\int\frac{4\text{x}-1}{2\text{x}^2-\text{x}}\text{ dx}-\frac{7}{16}\int\frac{1}{\text{x}^2-2\text{x}\big(\frac{1}{4}\big)+\big(\frac{1}{4}\big)^2-\big(\frac{1}{4}\big)^2}\text{ dx}$ $\text{I}=\frac{1}{8}\int\frac{4\text{x}-1}{2\text{x}^2-\text{x}}\text{ dx}-\frac{7}{16}\int\frac{1}{\big(\text{x}-\frac{1}{4}\big)^2-\big(\frac{1}{4}\big)^2}\text{ dx}$ $\text{I}=\frac{1}{8}\log\big|2\text{x}^2+\text{x}\big|-\frac{7}{16}\times\frac{1}{2\big(\frac{1}{24}\big)}\log\bigg|\frac{\text{x}-\frac{1}{4}-\frac{1}{4}}{\text{x}-\frac{1}{4}+\frac{1}{4}}\bigg|+\text{C}_2$ $\Big[\text{since},\int\frac{1}{\text{x}^2-\text{a}^2}\text{ dx}=\frac{1}{2\text{a}}\log\Big|\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\Big|+\text{C}\Big]$ $\text{I}=\frac{1}{8}\log|\text{x}|+\frac{1}{8}\log|2\text{x}-1|-\frac{7}{8}\log|1-2\text{x}|+\frac{7}{8}\log2+\frac{7}{8}\log|\text{x}|+\text{C}_2$ $\text{I}_1=\log\big|\text{x}\big|-\frac{3}{4}\log|1-2\text{x}|+\text{C}_3\ ....(2)$ $\Big[\text{Say},\text{C}_3=\text{C}_2+\frac{7}{8}\log2\Big]$ Using equation (1) and (2) $\text{I}=\frac{1}{2}\text{x}+\log|\text{x}|-\frac{3}{4}\log|1-2\text{x}|+\text{C}$

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