Question

If $\int \frac{2e^{5x}+e^{4x}-4e^{3x}+4e^{2x}+2e^x}{(e^{2x}+4){(e^{2x}-1)}^2}dx$ $={\tan }^{-1}\left(e^{x/2}\right)-\frac{K}{248(e^{2x}-1)}+C$ then K is equal to.

Answer 1

$I=\int \frac{2e^{5x}+e^{4x}-4e^{3x}+4e^{2x}+2e^x}{(e^{2x}+4){(e^{2x}-1)}^2}dx$

Put $e^x=t\Rightarrow e^{x}dx=dt$

$I=\int \frac{2t^4+t^3-4t^2+4t+2}{(t^2+4){(t^2-1)}^2}dt=\int \frac{2t^4-4t^2+2}{(t^2+4){(t^2-1)}^2}dt+\int \frac{t^3+4t}{(t^2+4){(t^2-1)}^2}dt$

$=2\int \frac{{(t^2-1)}^2}{{(t^2+4)(t^2-1)}^2}dt+\int \frac{t(t^2+4)}{(t^2+4){(t^2-1)}^2}dt$

$=2\int \frac{1}{(t^2+4)}dt+\frac{1}{2}\int \frac{2t}{{(t^2-1)}^2}dt={\tan }^1\frac{t}{2}-\frac{1}{2(t^2-1)}+c$

$I={\tan }^1\frac{e^x}{2}-\frac{1}{2(e^{2x}-1)}+c$

Comparing with, $\int \frac{2e^{5x}+e^{4x}-4e^{3x}+4e^{2x}+2e^x}{(e^{2x}+4){(e^{2x}-1)}^2}dx$ $={\tan }^{-1}\left(e^{x/2}\right)-\frac{K}{248(e^{2x}-1)}+C$

Therefore $K=124$