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Section 5.1 #41 : Calculate the derivative of $f(x)=\ln(3x)$.
Solution: Using the chain rule, we get
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}x} \ln(3x) &= \dfrac{1}{3x} \dfrac{\mathrm{d}}{\mathrm{d}x} [3x] \\
&= \dfrac{3}{3x} \\
&= \dfrac{1}{x}.
\end{array}$$
Section 5.1 #55: Calculate the derivative of $f(x)=\ln(\ln(x^2))$.
Solution: Using the chain rule twice, we get
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}x} \ln(\ln(x^2)) &= \dfrac{1}{\ln(x^2)} \dfrac{\mathrm{d}}{\mathrm{d}x} \ln(x^2) \\
&= \dfrac{1}{x^2 \ln(x^2)} \dfrac{\mathrm{d}}{\mathrm{d}x} x^2 \\
&= \dfrac{2x}{x^2 \ln(x^2)} \\
&= \dfrac{2}{x\ln(x^2)}.
\end{array}$$
Section 5.2 #14 : Calculate $\displaystyle\int \dfrac{x^2+4x}{x^3+6x^2+5} \mathrm{d}x$.
Solution: Let $u=x^3+6x^2+5$ so that $\mathrm{d}u=3x^2+12x \mathrm{d}x$. Rearrange this formula to get $\dfrac{1}{3} \mathrm{d}u = x^2+4x \mathrm{d}x$. Now using $u$-substitution, compute
$$\begin{array}{ll}
\displaystyle\int \dfrac{x^2+4x}{x^3+6x^2+5} \mathrm{d}x &= \dfrac{1}{3} \displaystyle\int \dfrac{1}{u} \mathrm{d}u \\
&= \dfrac{1}{3} \ln (|u|) + C \\
&= \dfrac{1}{3} \ln(|x^3+6x^2+5|) + C.
\end{array}$$
Section 5.2 #31 : Calculate $\displaystyle\int \cot \left( \dfrac{\theta}{3} \right) \mathrm{d}\theta$.
Solution: Recall that by definition, $\cot(x) = \dfrac{\cos(x)}{\sin(x)},$ and so $\cot \left( \dfrac{\theta}{3} \right) = \dfrac{\cos(\frac{\theta}{3})}{\sin(\frac{\theta}{3})}$. Let $u=\sin \left( \dfrac{\theta}{3} \right)$ so that $\mathrm{d}u = \dfrac{1}{3} \cos \left( \dfrac{\theta}{3} \right) \mathrm{d}x$. Rearrange this formula to get $3 \mathrm{d}u = \cos \left( \dfrac{\theta}{3} \right) \mathrm{d}\theta$. Now compute
$$\begin{array}{ll}
\displaystyle\int \cot \left( \dfrac{\theta}{3} \right) \mathrm{d}\theta &= \displaystyle\int \dfrac{\cos(\frac{\theta}{3})}{\sin(\frac{\theta}{3})} \mathrm{d}\theta \\
&= 3 \displaystyle\int \dfrac{1}{u} \mathrm{d}u \\
&= 3 \ln (|u|) + C \\
&= 3 \ln \left( \left| \sin \left( \dfrac{\theta}{3} \right) \right| \right) + C.
\end{array}$$
Section 5.2 #52 : Calculate $\displaystyle\int_e^{e^2} \dfrac{1}{x \ln(x)} \mathrm{d}x$.
Solution: Let $u=\ln(x)$ so that $\mathrm{d}u = \dfrac{1}{x} \mathrm{d}x$. Note that if $x=e$ then $u=\ln(e)=1$ and if $x=e^2$ then $u=\ln(e^2)=2$. Now compute (recalling that $\ln(1)=0$)
$$\begin{array}{ll}
\displaystyle\int_e^{e^2} \dfrac{1}{x\ln(x)} \mathrm{d}x &= \displaystyle\int_1^2 \dfrac{1}{u} \mathrm{d}u \\
&= \ln(|u|) \Bigg|_1^2 \\
&= \ln(|2|) - \ln(|2|) \\
&= \ln(2) - \ln(1) \\
&= \ln(2).
\end{array}$$
Section 5.2 #55 : Calculate $\displaystyle\int_1^2 \dfrac{1-\cos(\theta)}{\theta-\sin(\theta)} \mathrm{d}\theta$.
Solution: Let $u=\theta-\sin(\theta)$ so that $\mathrm{d}u=1-\cos(\theta) \mathrm{d}\theta$. If $\theta=1$ then $u=1-\sin(1)$ and if $\theta=2$ then $u=2-\sin(2)$. Now compute
$$\begin{array}{ll}
\displaystyle\int_1^2 \dfrac{1-\cos(\theta)}{\theta-\sin(\theta)} \mathrm{d}\theta &= \displaystyle\int_{1-\sin(1)}^{2-\sin(2)} \dfrac{1}{u} \mathrm{d}u \\
&= \ln (|u|) \Bigg|_{1-\sin(1)}^{2-\sin(2)} \\
&= \ln(|2-\sin(2)|) - \ln(|1-\sin(1)|).
\end{array}$$
