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(a) If we expand the integrand, one part of it is quite familiar–a portion of a circle. So, we split the specified integral in two.

$$\int_0^3(x+1)\sqrt{9-x^2} \ \mathrm{d}x = \int_0^3\sqrt{9-x^2} \ \mathrm{d}x +\int_0^3x\sqrt{9-x^2} \ \mathrm{d}x$$

The first piece represents the area above the $x$–axis and below the curve $y=\sqrt{9-x^2}$, i.e. $x^2+y^2=9$, with $0\le x\le 3$. That's the area of one quadrant of a disk of radius $3$. So

$$\int_0^3\sqrt{9-x^2} \ \mathrm{d}x =\frac{1}{4}(\pi \cdot 3^2) = \frac{9}{4}\pi$$

For the second part, we substitute $u=9-x^2$, $\mathrm{d}u= -2x\,\mathrm{d}x$. Note $u(0)=9$ and $u(3)=0$. So,

$$\begin{aligned} \int_0^3x\sqrt{9-x^2} \ \mathrm{d}x = \int_9^0\sqrt{u} \ \frac{\mathrm{d}u}{-2} =-\frac{1}{2}\left[\frac{u^{3/2}}{3/2}\right]_9^0 =-\frac{1}{2}\left[-\frac{27}{3/2}\right] =9 \end{aligned}$$

All together,

$$\int_0^3(x+1)\sqrt{9-x^2} \ \mathrm{d}x = \frac{9}{4}\pi + 9$$

(b) The integrand is of the form $N(x)/D(x)$ with $D(x)$ already factored and $N(x)$ of lower degree. We immediately look for a partial fractions decomposition:

$$\frac{4x+8}{(x-2)(x^2+4)} = \frac{A}{x-2} + \frac{Bx+C}{x^2+4}.$$

Multiplying through by the denominator yields

$$\begin{aligned} 4x+8 &= A(x^2+4) + (Bx+C)(x-2) \tag{$*$} \end{aligned}$$

Setting $x=2$ we find:

$$\begin{aligned} 8 +8 = A(4 + 4) + 0 \implies 16 =8A \implies A = 2 \end{aligned}$$

Substituting $A=2$ in $(*)$ gives

$$\begin{aligned} &&4x+8 &= A(x^2+4) + (Bx+C)(x-2) \\ &\implies & -2x^2+4 x &=(x-2)(Bx + C) \\ &\implies\qquad & (-2x)(x-2)&=(Bx + C)(x-2) \\ & \implies& B& =-2,~C=0 \end{aligned}$$

So we have found that $A=2$, $B=-2$, and $C=0$. Therefore

$$\begin{aligned} \int\frac{4x+8}{(x-2)(x^2+4)}\ \mathrm{d}x &= \int \bigg( \frac{{red}2}{x-2} {red}- \frac{{red}2x}{x^2+4} \bigg) \,\mathrm{d}x \\ &= 2 \log |x-2| - \log (x^2 + 4) + C \end{aligned}$$

Here the second integral was found just by guessing an antiderivative. Alternatively, one could use the substitution $u=x^2+4$, $\mathrm{d}u=2x\,\mathrm{d}x$.

(c) The given integral is improper, but only because of its infinite limits of integration. (The integrand is continuous for all real numbers.) So, we'll have to take two limits. Before we do that, though, let's find the antiderivative. We would like to use the substitution $u=e^x$, $\mathrm{d}u=e^x~\mathrm{d}x$. That is, $\frac{1}{u}~\mathrm{d}u = \mathrm{d}x$.

$$\begin{aligned} \int \frac{1}{e^x+e^{-x}}~\mathrm{d}x&=\int \frac{1}{u(u+\frac{1}{u})}~\mathrm{d}u = \int\frac{1}{u^2+1} = \arctan u+C \\&=\arctan(e^x)+C \\ \text{Now we can deal with the limits of integration.} \\ \int _{ -\infty}^\infty\frac{1}{e^x+e^{-x}}~\mathrm{d}x&= \int _{ -\infty}^0\frac{1}{e^x+e^{-x}}~\mathrm{d}x+ \int _{ 0}^\infty\frac{1}{e^x+e^{-x}}~\mathrm{d}x\\&= \lim_{a \to -\infty}\left[\int_{a}^0\frac{1}{e^x+e^{-x}}~\mathrm{d}x\right]+ \lim_{b \to \infty}\left[\int_{ 0}^b\frac{1}{e^x+e^{-x}}~\mathrm{d}x\right]\\&= \lim_{a \to -\infty}\big[\arctan(e^x)\big]_a^0+ \lim_{b \to \infty}\big[\arctan(e^x)\big]_0^b\\&= \lim_{a \to -\infty}\big[\arctan(e^0)-\arctan(e^a)\big]+ \lim_{b \to \infty}\big[\arctan(e^b)-\arctan(e^0)\big]\\&= \lim_{a \to -\infty}\big[-\arctan(e^a)\big]+ \lim_{b \to \infty}\big[\arctan(e^b)\big]\\ &=-\arctan(0)+\frac{\pi}{2} = \frac{\pi}{2} \end{aligned}$$