Question

$\int \frac{1-x^2}{x(1-2x)}dx=$

Answer 1

$\int \frac{1-x^2}{\left(1-2x\right)x}dx$

$=\int \frac{x^2-1}{x\left(2x-1\right)}dx$

$\begin{array}{c}=\int \left(\frac{x-2}{2x\left(2x-1\right)}+\frac{1}{2}\right)dx\\ \\ =\frac{1}{2}\int \frac{x-2}{x\left(2x-1\right)}dx+\frac{1}{2}\int 1dx---\left(1\right)\\ \\ \int \frac{x-2}{x\left(2x-1\right)}dx\\ \\ =\int \left(\frac{2}{x}-\frac{3}{2x-1}\right)dx\\ \\ =2\int \frac{1}{x}dx-3\int \frac{1}{2x-1}dx---\left(2\right)\\ \\ \int \frac{1}{2x-1}dx\end{array}$

Now substitute

$u=2x-1$

$dx=\frac{1}{2}du$

$\begin{array}{c}=\frac{1}{2}\int \frac{1}{u}du\\ \\ =\frac{\ln \left(u\right)}{2}\\ \\ =\frac{\ln \left(2x-1\right)}{2}\\ \\ 2\int \frac{1}{x}dx-3\int \frac{1}{2x-1}dx\\ \\ =2\ln \left(x\right)-\frac{3\ln \left(2x-1\right)}{2}\\ \\ \frac{1}{2}\int \frac{x-2}{x\left(2x-1\right)}dx+\frac{1}{2}\int 1dx\\ \\ =-\frac{3\ln \left(\left|2x-1\right|\right)}{4}+\ln \left(\left|x\right|\right)+\frac{x}{2}+C\end{array}$