Evaluate the following integrals: (1-x^2) x(1-2x) dx

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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)

$\text{I}=\frac{1}{2}\text{x}+\log|\text{x}|-\frac{3}{4}\log|1-2\text{x}|+\text{C}$