$$\int\frac{dx}{1-\cos3x}dx = -\frac{1}{3}(\cot(3x)+\csc(3x)) + C$$
$\frac{1}{1-\cos(3x)} = \frac{1+ \cos(3x)}{\sin^2(3x)}$
$\frac{1}{1-\cos(3x)} = \csc^2(3x) + \frac{\cos(3x)}{\sin^2(3x)}$
$\int\frac{dx}{1-\cos(3x)}dx = \frac{1}{3}\left(\,\int\csc^{2}(3x)d(3x) +\int\sin^{-2}(3x)d\sin(3x)\right) $
$\int\frac{dx}{1-\cos(3x)}dx = -\frac{1}{3}(\cot(3x)+\csc(3x)) + C$
$$\int sin(7x)cos(5x)dx = -\frac{1}{2}\left(\,\frac{\cos(2x)}{2}+\frac{\cos(12x)}{12}\right) + C$$
$\sin{A}\cos{B} = \frac{1}{2}\left(\,\sin(A+B)+\sin(A-B)\right) $
$\sin(7x)\cos(5x) = \frac{1}{2}\left(\,\sin(12x)+\sin(2x)\right)$
$\int sin(7x)cos(5x)dx = \int \frac{1}{2}\left(\,\sin(12x)+\sin(2x)\right)dx$
$\int sin(7x)cos(5x)dx = -\frac{1}{2}\left(\,\frac{\cos(2x)}{2}+\frac{\cos(12x)}{12}\right) + C$
$$\int\frac{e^x}{\sqrt{1-e^{2x}}}dx = \sin^{-1}(e^x) + C$$
$u = e^x$
$\frac{du}{e^x} = dx$
$\int\frac{e^x}{\sqrt{1-e^{2x}}}dx = \int\frac{e^x}{\sqrt{1-u^2 }}\frac{du}{e^x}$
$\int\frac{e^x}{\sqrt{1-e^{2x}}}dx = \int\frac{1}{\sqrt{1-u^2 }}du$
$\int\frac{1}{\sqrt{a^2-u^2}}du = \frac{1}{a}\sin^{-1}\left(\,\frac{u}{a}\right) + C$
$\int\frac{e^x}{\sqrt{1-e^{2x}}}dx = \sin^{-1}(e^x) + C$
$$\int\frac{dx}{\sqrt{x}(1+x)} = 2\tan^{-1}\left(\,\sqrt{x}\right) + C$$
$u = \sqrt{x}$
$2\sqrt{x}du = dx$
$x = u^2$
$\int\frac{dx}{\sqrt{x}(1+x)} = \int\frac{1}{u(1+u^2)}2udu$
$\int\frac{dx}{\sqrt{x}(1+x)} = 2\int\frac{1}{(1+u^2)}du$
$\int\frac{1}{(a^2+u^2)}du = \frac{1}{a}\tan^{-1}\left(\,\frac{u}{a}\right) + C$
$\int\frac{dx}{\sqrt{x}(1+x)} = 2\tan^{-1}\left(\,\sqrt{x}\right) + C$
$$\int\frac{2x^3-6x^2+8x}{x^2+4}dx = x^2 - 6x + 12\tan^{-1}\frac{x}{2} + C$$
$\frac{2x^3-6x^2+8x}{x^2+4} = 2x - 6 + \frac{24}{x^2 + 4}$
$\int\frac{2x^3-6x^2+8x}{x^2+4}dx = \int\left(\,2x - 6 + \frac{24}{x^2 + 4}\right)dx $
$\int\frac{2x^3-6x^2+8x}{x^2+4}dx = \int2xdx - \int6dx + 24\int\frac{1}{2^2 + x^2}dx $
$\int\frac{1}{(a^2+u^2)}du = \frac{1}{a}\tan^{-1}\left(\,\frac{u}{a}\right) + C$
$\int\frac{2x^3-6x^2+8x}{x^2+4}dx = x^2 - 6x + 12\tan^{-1}\frac{x}{2} + C$
$$\int\frac{2-x}{4x^2+4x-3}dx = -\frac{7}{16}\ln\left|\,2x+3\right|+\frac{3}{16}\ln\left|\,2x-1\right|+C $$
$\frac{2-x}{4x^2+4x-3} = \frac{2-x}{(2x+3)(2x-1)}$
$\frac{2-x}{(2x+3)(2x-1)} = -\frac{7}{8}\left(\,\frac{1}{2x+3}\right) + \frac{3}{8}\left(\,\frac{1}{2x-1}\right)$
$\int\frac{2-x}{4x^2+4x-3}dx = \int\left(\,-\frac{7}{8}\left(\,\frac{1}{2x+3}\right) + \frac{3}{8}\left(\,\frac{1}{2x-1}\right)\right)dx $
$\int\frac{2-x}{4x^2+4x-3}dx = -\frac{7}{16}\ln\left|\,2x+3\right|+\frac{3}{16}\ln\left|\,2x-1\right|+C $
$$\int\frac{3x^4+3x^3-5x^2+x-1}{x^2+x-2}dx = x^3 + x + \frac{1}{3}\ln\left|\,\frac{x-1}{x+2}\right|+ C $$
$\frac{3x^4+3x^3-5x^2+x-1}{x^2+x-2} = 3x^2 + 1 + \frac{1}{(x+2)(x-1)}$
$\frac{1}{(x+2)(x-1)} = -\frac{1}{3}\left(\,\frac{1}{x+2}\right) + \frac{1}{3}\left(\,\frac{1}{x-1}\right) $
$\int\frac{3x^4+3x^3-5x^2+x-1}{x^2+x-2}dx = \int\left(\,3x^2 +1 -\frac{1}{3}\left(\,\frac{1}{x+2}\right) + \frac{1}{3}\left(\,\frac{1}{x-1}\right)\right)dx $
$\int\frac{3x^4+3x^3-5x^2+x-1}{x^2+x-2}dx = x^3 + x - \frac{1}{3}\ln\left|\,x+2\right| + \frac{1}{3}\ln\left|\,x-1\right| + C$
$\int\frac{3x^4+3x^3-5x^2+x-1}{x^2+x-2}dx = x^3 + x + \frac{1}{3}\ln\left|\,\frac{x-1}{x+2}\right|+ C$
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