$$\int \log (1+x)^{1+x} \mathrm{~d} x=$$

$$\int\left(\frac{x+2}{x+4}\right)^2 \cdot e^x \mathrm{~d} x=$$

$\int \frac{\mathrm{d} x}{3-2 \cos 2 x}=\frac{\tan ^{-1}(\mathrm{f}(x))}{\sqrt{5}}+\mathrm{c}$, (where c is a constant of integration), then $f(\pi / 4)$ has the value

The value of $$\int \mathrm{e}^x\left(\frac{x^2+4 x+4}{(x+4)^2}\right) \mathrm{d} x$$ is :