Krzychau1

2021-12-31

Show that ${\int }_{1}^{2}\frac{{e}^{x}\left(x-1\right)}{x\left(x+{e}^{x}\right)}dx=\mathrm{ln}\left(\frac{2+{e}^{2}}{2+2e}\right)$ and ${\int }_{0}^{\frac{\pi }{2}}\frac{x\mathrm{sin}\left(2x\right)}{1+{\mathrm{cos}\left(2x\right)}^{2}}dx=\frac{{\pi }^{2}}{16}$

karton

Now add them up to get:
$2ȷ=\frac{\pi }{2}{\int }_{0}^{\frac{\pi }{2}}\frac{\mathrm{sin}\left(2x\right)}{1+{\mathrm{cos}}^{2}\left(2x\right)}dx⇒ȷ=-\frac{\pi }{8}\mathrm{arctan}$
$\left(\mathrm{cos}\left(2x\right)\right){|}_{0}^{\frac{\pi }{2}}=\frac{{\pi }^{2}}{16}$

nick1337

For the second integral, use ${\int }_{a}^{b}f\left(x\right)={\int }_{a}^{b}f\left(a+b-x\right)$ and add these together to get $I={\int }_{0}^{\frac{\pi }{2}}\frac{\frac{\pi }{4}\mathrm{sin}\left(2x\right)}{1+\mathrm{cos}\left(2x{\right)}^{2}}$. Substituting with $\mathrm{cos}\left(2x\right)=t$ gives the solution easily.

user_27qwe

Hint:
$\int \frac{2\mathrm{sin}\left(2x\right)}{1+{\mathrm{cos}}^{2}\left(2x\right)}dx=-\mathrm{arctan}\left(\mathrm{cos}\left(2x\right)\right)$

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