As stated in the title, I am currently working on solving $$\int \cos(5x)\sinh(6x)\,dx.$$
After using integration by parts twice, I have gotten to the point where I have $$\frac{\sinh(6x)\sin(5x)}{5}+\frac{6\cosh(6x)\cos(5x)}{25}+\int \frac{-36\cos(5x)\sinh(6x)}{25}.$$
I've rewritten the first part by multiplying the first term by $5$, giving $$\frac{5\sinh(6x)\sin(5x)+6\cosh(6x)\cos(5x)}{25}+\int \frac{-36\cos(5x)\sinh(6x)}{25}.$$
I can see that the new integral is a multiple of the original, and using an online calculator I am told the solution is $$-\frac{-5\sinh(6x)\sin(5x)-6\cosh(6x)\cos(5x)}{61}+C.$$
However I can't grasp how this answer was calculated from where I am currently, though I think the $61$ in the denominator is throwing me off.
Let $I$ denote your original integral. You've worked out that $$ I = \frac{\sinh(6x)\sin(5x)}{5}+\frac{6\cosh(6x)\cos(5x)}{25}+\int \frac{-36\cos(5x)\sinh(6x)}{25} \\ = \frac{\sinh(6x)\sin(5x)}{5}+\frac{6\cosh(6x)\cos(5x)}{25}+ \frac{-36}{25}I. $$ From this, you know that $$ (1 + 36/25) I = \frac{\sinh(6x)\sin(5x)}{5}+\frac{6\cosh(6x)\cos(5x)}{25} $$ and you can now solve for $I$.