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Section 5.4 #49: Calculate $\dfrac{\mathrm{d}}{\mathrm{d}x} \dfrac{e^x+1}{e^x-1}$.
Solution: Using the quotient rule, compute
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}x} \dfrac{e^x+1}{e^x-1} &= \dfrac{(e^x-1)(e^x) - (e^x+1)(e^x)}{(e^x-1)^2} \\
&=\dfrac{e^x[(e^x-1)-(e^x+1)]}{(e^x-1)^2} \\
&=\dfrac{-2e^x}{(e^x-1)^2}. \blacksquare
\end{array}$$
Section 5.4 #52: Calculate $\dfrac{\mathrm{d}}{\mathrm{d}x} e^{2x}\tan(2x)$.
Solution: Recall that $\dfrac{\mathrm{d}}{\mathrm{d}x} \tan(x)=\sec^2(x)$. Therefore use the product rule to compute
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}x} e^{2x}\tan(2x) &= \dfrac{\mathrm{d}}{\mathrm{d}x}[e^{2x}] \tan(2x) + e^{2x} \dfrac{\mathrm{d}}{\mathrm{d}x} \tan(2x) \\
&= 2e^{2x}\tan(2x) +2e^{2x}\sec^2(2x). \blacksquare
\end{array}$$
Section 5.4 #106: Calculate $\displaystyle\int \dfrac{e^{2x}+2e^x+1}{e^x} \mathrm{d}x$.
Solution: You could make the substitution $u=e^x$, but it is easier to use algebra to simplify the integrand:
$$\dfrac{e^{2x}+2e^x+1}{e^x} = e^x+2+e^{-x},$$
so we may write
$$\displaystyle\int \dfrac{e^{2x}+2e^x+1}{e^x} \mathrm{d}x = \displaystyle\int e^x + 2 + e^{-x} \mathrm{d}x = \displaystyle\int e^x \mathrm{d}x + \displaystyle\int 2 \mathrm{d}x + \displaystyle\int e^{-x} \mathrm{d}x.$$
The first two of these integrals are straightforward, and for the third one we make the substitution $u=-x$ so that $-\mathrm{d}u=\mathrm{d}x$ and we compute
$$\begin{array}{ll}
\displaystyle\int \dfrac{e^{2x}+2e^x+1}{e^x} \mathrm{d}x &= \displaystyle\int e^x \mathrm{d}x + \displaystyle\int 2 \mathrm{d}x + \displaystyle\int e^{-x} \mathrm{d}x \\
&= e^x + 2x - \displaystyle\int e^u \mathrm{d}u \\
&= e^x + 2x - e^u + C \\
&= e^x + 2x - e^{-x} + C. \blacksquare
\end{array}$$
Section 5.4 #107: Calculate $\displaystyle\int e^{-x} \tan(e^{-x}) \mathrm{d}x$.
Solution: First we recall how to compute $\displaystyle\int \tan(x) \mathrm{d}x$: since $\tan(x) = \dfrac{\sin(x)}{\cos(x)}$ we make the substitution $u=\cos(x)$ so that $-\mathrm{d}u=\sin(x)\mathrm{d}x$ and we see
$$\begin{array}{ll}
\displaystyle\int \tan(x) \mathrm{d}x &= \displaystyle\int \dfrac{\sin(x)}{\cos(x)} \mathrm{d}x \\
&=-\displaystyle\int \dfrac{1}{u} \mathrm{d}u \\
&= -\log(u) + C \\
&= -\log(\cos(x)) + C.
\end{array}$$
Now we may proceed with the original problem. To do it, make the subtitution $w=e^{-x}$ so that $-\mathrm{d}w = e^{-x} \mathrm{d}x$ (note: I'm using $w$ instead of $u$ since I already used $u$ above). Now compute
$$\begin{array}{ll}
\displaystyle\int e^{-x} \tan(e^{-x}) \mathrm{d}x &= -\displaystyle\int \tan(w) \mathrm{d}w \\
&= -[-\log(\cos(w))] + C \\
&= \log(\cos(e^{-x})) + C. \blacksquare
\end{array}$$
Section 5.5 #43: Calculate $\dfrac{\mathrm{d}}{\mathrm{d}t} t^2 2^t$.
Solution: Recall that we may write
$$2^t = e^{\log(2^t)} = e^{t \log(2)}.$$
Now use the product rule to compute
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}t} t^2 2^t &= \dfrac{\mathrm{d}}{\mathrm{d}t} [t^2] 2^t + t^2 \dfrac{\mathrm{d}}{\mathrm{d}t} e^{t\log(2)} \\
&= (2t)(2^t) + (t^2)(\log(2)e^{t\log(2)}) \\
&= t2^t \left[ 2 + t\log(2) \right]. \blacksquare
\end{array}$$
Section 5.5 #81: Calculate $\displaystyle\int_0^1 5^x-3^x \mathrm{d}x$.
Solution: Recall how to compute $\displaystyle\int a^x \mathrm{d}x$ for any $a>0$: first rewrite $a^x=e^{\log(a^x)} = e^{x\log(a)}$ and use the $u$-substitution $u=x\log(a)$ to get $\dfrac{1}{\log(a)} \mathrm{d}u = \mathrm{d}x$. Then,
$$\begin{array}{ll}
\displaystyle\int a^x \mathrm{d}x &= \displaystyle\int e^{x\log(a)} \mathrm{d}x \\
&= \dfrac{1}{\log(a)} \displaystyle\int e^u \mathrm{d}u \\
&= \dfrac{1}{\log(a)}e^u + C \\
&= \dfrac{1}{\log(a)} e^{x\log(a)} + C \\
&= \dfrac{1}{\log(a)} a^x + C.
\end{array}$$
Applying this formula for $a=5$ and for $a=3$ allows us to compute
$$\begin{array}{ll}
\displaystyle\int_0^1 5^x-3^x \mathrm{d}x &= \displaystyle\int_0^1 5^x \mathrm{d}x - \displaystyle\int_0^1 3^x \mathrm{d}x \\
&= \Bigg[ \dfrac{1}{\log(5)} 5^x \Bigg|_0^1 - \Bigg[ \dfrac{1}{\log(3)} 3^x \Bigg|_0^1 \\
&= \dfrac{5-1}{\log(5)} - \dfrac{3-1}{\log(3)} \\
&= \dfrac{4}{\log(5)} - \dfrac{2}{\log(3)}. \blacksquare
\end{array}$$
