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I have the following integral:

$$\int \left(\frac{e^{3x}+1}{e^{x}+1}\right)\frac{\mathrm{d} }{\mathrm{d} x}$$

I divided him into 2 parts:

First is: $\int (\frac{e^{3x}}{e^{x}+1})\frac{\mathrm{d} }{\mathrm{d} x}$

And Second:$\int (\frac{1}{e^{x}+1})\frac{\mathrm{d} }{\mathrm{d} x}$

I need to use the reverse chain rule(on the first one, the second I think I am

able to solve) but I can't link the things that it will form an integral of this

kind:

$\int uv'=uv-\int u'v$

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    $\begingroup$ That's product rule in reverse ;) $\endgroup$ – Kaj Hansen Jun 21 '18 at 20:22
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$$e^{3x}+1=(e^x+1)(e^{2x}-e^x+1) $$

the integral becomes

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

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$$\frac{e^{3x}+1}{e^x+1}=e^{2x}-e^x+1$$

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Hint.

$$ \frac{e^{3x}+1}{e^x+1} = e^{2x}-e^x+1 $$

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