Given is the following function:
$f(x):=\sinh(2x)\sin(4x)$
With partial integration I found following antiderivative:
$\int\!\sinh(2x)\sin(4x)\mathrm{d}x$ $=\frac{1}{2}\cosh(2x)\sin(4x)-\int\!\frac{1}{2}\cosh(2x)4\cos(4x)$ $=\frac{1}{2}\cosh(2x)\sin(4x)-2\int\!\cosh(2x)\cos(4x)$ $=\frac{1}{2}\cosh(2x)\sin(4x)-2\frac{1}{2}\sinh(2x)\cos(4x)-2\int\!-\frac{1}{2}\sinh(2x)4\sin(4x)$ $=\frac{1}{2}\cosh(2x)\sin(4x)-\sinh(2x)\cos(4x)+4\int\!\sinh(2x)\sin(4x)$ $=-\frac{1}{3}(\frac{1}{2}\cosh(2x)\sin(4x)-\sinh(2x)\cos(4x))$
Nevertheless WolframAlpha has found another solution.
What did I do wrong?