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After trying this multiple ways, I give up. Here's the integral:

$$\int e^{-x}\arctan(e^x)\,dx$$

I have set

$u=\arctan(e^x)$ and $dv=e^{-x}d\,x$

and have obtained $du=\dfrac{e^x \, dx}{1+e^{2x}}$ and $v=-e^{-x}$

Using the integration by parts formula,

$$\int u\, dv = uv - \int v\,du$$

I got $-e^{-x}\arctan(e^x)+\int \dfrac{1}{1+e^{2x}}\,dx$

How would I solve $\int \dfrac{1}{1+e^{2x}}\,dx$ ? That's the part I'm stuck on.

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up vote 3 down vote accepted

$$\int \frac{dx}{1+e^{2 x}} = \int dx \frac{e^{-x}}{e^x+e^{-x}} = - \int du \frac{1}{u+u^{-1}}$$

which is

$$-\int du \frac{u}{1+u^2} = -\frac12 \log{(1+u^2)} + C = -\frac12 \log{\left (1+e^{-2 x}\right)} + C$$

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Everyone's responses were very helpful, and after some further tinkering, I'd like to add my own.

We can pull out the $e^{2x}$ in the denominator of

$$\int \dfrac{1}{1+e^{2x}}\,dx$$

and obtain $$\int \dfrac{1}{e^{2x}(1+e^{-2x})}\,dx$$

which simplifies to $$\int \dfrac{e^{-2x}}{1+e^{-2x}}\,dx.$$

We can now use u-substitution by setting $$u=1+e^{-2x}$$

and, by taking the derivative of $u$, we get $$du=-2e^{-2x}dx.$$

The integral now becomes $$-\dfrac{1}{2}\int\dfrac{du}{u}$$

which yields $$-\frac12\ln{(1+e^{-2x})}+C.$$

Once again, thank you all for your kind help. It was greatly appreciated.

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HINT: $$\frac{1}{1+e^{2x}}=1-\frac{e^{2x}}{1+e^{2x}}$$ and substitute $u = e^{2x}$.

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In $I = \int e^{-x}\arctan(e^x)\,dx$, let $y = e^{-x}$.

Since $dy = -e^{-x} dx$, $I = \int -dy \arctan(1/y) =-\int \arctan(\pi/2-y)\, dy =\int \arctan(y-\pi/2)\, dy $

and then use $\int \arctan x\, dx = x \arctan x - (1/2) ln(1+x^2) + C $.

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