Let I = ∫ $e^x\left(\frac{\sin 4x-4}{1-\cos 4x}\right)$ dx = ∫ $e^x\left(\frac{2\sin 2x \cos 2x - 4}{2\sin^2(2x)}\right)$ dx [Using, sin2x 2sinx . cosx and 2 $\sin^2$ x = 1 - cos(2x) = ∫ $e^x\left(\frac{2(\sin(2x)\cos(2x)-4)}{2\sin^2 x}\right)$ dx = ∫ $e^x\left(\frac{\sin(2x)\cos(2x)}{\sin^2 2x} - \frac{2}{\sin^2 2x}\right)$ dx = ∫ $e^x$ 9cot (2x) - 2 $\csc^2$ 2x) dx Now, let f(x) = cot (2x) then f’(x) = -2 $\csc^2$ 2x I = ∫ $e^x$ (f (x) + f' (x)) dx So, I = $e^x$ f(x) + C = $e^x$ cot 2x + C , where C is a constant Therefore, $\int e^x\left(\frac{\sin 4x-4}{1-\cos 4x}\right)$ dx = $e^x$ cot (2x) + C OR ∫ $\frac{1-x^2}{x(1-2x)}$ dx Here $\frac{1- x^2}{x(1-2x)}$ is an improper rational fraction Reducing it to proper rational fraction gives $\frac{1-x^2}{x(1-2x)}$ = $\frac{1}{2} + \frac{1}{2}(\frac{2-x}{x(1-2x)}$ ... (i) Now, let $\frac{2-x}{x(1-2x)}$ = $\frac{A}{x} + \frac{B}{1-2x}$ ⇒ $\frac{2-x}{x(1-2x)}$ = $\frac{A(1-2x)+Bx}{x(1-2x)}$ ⇒ 2 - x = A - x (2A - B) Equating the coefficients we get ,A = 2 and B = 3 So, $\frac{2-x}{x(1-2x)}$ = $\frac{2}{x} + \frac{3}{1-2x}$ Substituting in equation (1), we get $\frac{1-x^2}{x(1-2x)}$ = $\frac{1}{2} + \frac{1}{2}\left(\frac{2}{x} + \frac{3}{1-2x}\right)$ i.e. ∫ $\frac{1-x^2}{x(1-2x)}$ dx = ∫ $\left[\frac{1}{2} + \frac{1}{2}\left(\frac{2}{x} + \frac{3}{1-2x}\right)\right]$ dx = ∫ $\frac{dx}{2}$ + ∫ $\frac{dx}{2}$ + $\frac{3}{2}\int\frac{dx}{1-2x}$ = $\frac{x}{2}$ + log |x| + $\frac{3}{2} \times \frac{1}{-2}$ log |1 - 2x| + C = $\frac{x}{2}$ + log |x| - $\frac{3}{4}$ log |1 - 2x| + C