#### Provide Solution For R.D. Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.20 Question 3 Maths Textbook Solution.

Answer: $\frac{x}{2}+\log |x|-\frac{3}{4} \log |2 x-1|+c$

Given: $\int \frac{\left(1-x^{2}\right)}{x(1-2 x)} d x$

Hint: Using Partial fraction

Explanation:

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

$I=\int \frac{x^{2}-1}{2 x^{2}-x} d x$

$\frac{x^{2}-1}{2 x^{2}-x}=\frac{1}{2}+\frac{\frac{1}{2} x-1}{2 x^{2}-x}$

$\therefore \int \frac{x^{2}-1}{2 x^{2}-x} d x=\int \frac{1}{2} d x+\int \frac{\frac{x}{2}-1}{2 x^{2}-x} d x$

$=\frac{1}{2} x+I_{1}$      ............(1) Where $I_{1}=\int \frac{\frac{x}{2}-1}{2 x^{2}-x} d x$

$\frac{\frac{x}{2}-1}{2 x^{2}-x}=\frac{\frac{x}{2}-1}{x(2 x-1)}=\frac{A}{x}+\frac{B}{2 x-1}$

Multiplying by $x(2 x-1)$

$\frac{x}{2}-1=A(2 x-1)+B(x)$

Putting x = 0

$-1=A(-1)+B(0) \Rightarrow A=1$

Putting x = 2

\begin{aligned} &1-1=1(4-1)+B(2) \\ &0=3+2 B \\ &B=\frac{-3}{2} \end{aligned}

\begin{aligned} &\therefore \frac{\frac{x}{2}-1}{2 x^{2}-x}=\frac{1}{x}-\frac{3}{2(2 x-1)} \\ &\therefore I_{1}=\int \frac{1}{x} d x-\frac{3}{2} \int \frac{2}{2(2 x-1)} d x \\ &=\int \frac{1}{x} d x-\frac{3}{4} \int \frac{2}{(2 x-1)} d x \\ &=\log |x|-\frac{3}{4} \log |2 x-1| \end{aligned}

Put in (1)

$\frac{1}{2} x+\log |x|-\frac{3}{4} \log |2 x-1|+c$