Integrate the given function. \[ \frac{15 e^{5 r} d r}{e^{5 r}+2} \] \( \int \frac{15 e^{5 r} d r}{e^{5 r}+2}=\square \)

\( \frac{d}{dx}(\ln (\arctan (x+1))-\ln (\arctan (x))) \)

2. find Maclurin expansion of (2 Points) * \[ \begin{array}{l}f(x)=\sinh x \\ f(x)=1+\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\ldots \ldots \\ f(x)=x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\ldots \ldots \\ f(x)=x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\ldots \ldots \\ f(x)=x+\frac{x^{3}}{3!}+\frac{x^{5}}{5!}+\ldots\end{array} \]

\( f ( x ) = \int | x | d x \)

A container in the shape of an inverted cone of radius 3 metres and vertical height 4.5 metres is initially filled with liquid fertiliser. This fertiliser is released through a hole in the bottom of the container at a rate of \( 0.01 \mathrm{~m}^{3} \) per second. At time \( t \) seconds the fertiliser remaining in the container forms an inverted cone of height \( h \) metres. [The volume of a cone is \( V=\frac{1}{3} \pi r^{2} h \).] (i) Show that \( h^{2} \frac{\mathrm{~d} h}{\mathrm{~d} t}=-\frac{9}{400 \pi} \). (ii) Express \( h \) in terms of \( t \). (iii) Find the time it takes to empty the container, giving your answer to the nearest minute.

a) \( \iint_{\boldsymbol{R}}(\mathbf{5}-\boldsymbol{x}-\mathbf{2 y}) \boldsymbol{d} x \boldsymbol{d y} \) b) \( \iint_{\boldsymbol{R}}(\mathbf{5}-\boldsymbol{x}-\mathbf{2 y}) \boldsymbol{d y} \boldsymbol{d x} \) \[ R=\{(x, y) \mid 1 \leq x \leq 2,0 \leq y \leq 3\} \]