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