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:


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

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




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

Now, I integrate by parts:

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

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

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

$s=\frac 1t$

$t=\frac 1s$


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:



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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.