# Integration of $\int \frac{\arcsin{e^x}}{e^x}dx$

I've got a problem with this integral:

$$\int \frac{\arcsin{e^x}}{e^x}dx$$

I got such a result: $$\int\frac{\arcsin{e^x}}{e^x}dx=-\frac{\arcsin{e^x}}{e^x}-\ln|\sqrt{e^{-2x}-1}+e^{-x}|+C$$ but the wolphram alpha and the book where this integral is as an exersise give this answer:

$$\int\frac{\arcsin{e^x}}{e^x}dx=x-e^x\arcsin{e^x}-\ln(1+\sqrt{1-e^{2x}})+C$$

Where do I make a misteake? This is my solution:

$$\int \frac{\arcsin{e^x}}{e^x}dx=\int \frac{e^x\arcsin{e^x}}{e^{2x}}dx$$

Now,

$t=e^x$

$dt=e^xdx$

$$\int \frac{\arcsin{e^x}}{e^x}dx=\int\frac{\arcsin{t}}{t^2}dt$$

Now, I integrate by parts:

$u=\arcsin{t},\ v^{'}=\frac{1}{t^2}$

$u^{'}=\frac{1}{\sqrt{1-t^2}},\ v=-\frac 1t$

Hence, $$\int \frac{\arcsin{t}}{t^2}dt=-\frac{\arcsin{t}}{t}+\int\frac{dt}{t\sqrt{1-t^2}}$$

$s=\frac 1t$

$t=\frac 1s$

$dt=-\frac{1}{s^2}ds$

and I get

$$\int\frac{dt}{t\sqrt{1-t^2}}=-\int\frac{ds}{s^2\sqrt{1-\frac{1}{s^2}}\cdot \frac 1s}= -\int\frac{ds}{\sqrt{s^2-1}}=-\ln|\sqrt{s^2-1}+s|+C^{'}$$

From this we get:

$$\int\frac{\arcsin{e^x}}{e^x}dx=-\frac{\arcsin{e^x}}{e^x}-\ln|\sqrt{e^{-2x}-1}+e^{-x}|+C$$

Both are right. Note $$\ln | \sqrt{e^{-2x} -1} + e^{-x} | = \ln |e^{-x} (\sqrt{1 - e^{-2x}} + 1 ) | = \ln |e^{-x}| + \ln | \sqrt{1 - e^{-2x}} + 1 | = -x + \ln (\sqrt{1 - e^{-2x}} + 1 )$$ where the absolute values are unnecessary in the second and third steps due to the things being non-negative