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Section 5.1 #44) Compute $\dfrac{\mathrm{d}}{\mathrm{d}x} \ln(2x^2+1)$.
Solution: Compute
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}x} \ln(2x^2+1) &= \dfrac{1}{2x^2+1} \dfrac{\mathrm{d}}{\mathrm{d}x} \left[ 2x^2+1 \right] \\
&=\dfrac{4x}{2x^2+1}. \blacksquare
\end{array}$$
Section 5.1 #57) Compute $\dfrac{\mathrm{d}}{\mathrm{d}x} \ln \left( \sqrt{ \dfrac{x+1}{x-1} } \right)$.
Solution: A more difficult (but direct) method than I will use would be to directly compute
$$\dfrac{\mathrm{d}}{\mathrm{d}x} \ln \left( \sqrt{ \dfrac{x+1}{x-1} } \right)=\dfrac{1}{\sqrt{\frac{x+1}{x-1}}} \dfrac{\mathrm{d}}{\mathrm{d}x} \left[ \sqrt{ \dfrac{x+1}{x-1} } \right],$$
which would then require a second chain rule involving the quotient rule. The easier method for this antiderivative is to rewrite the $\sqrt{}$ symbol using exponent $\dfrac{1}{2}$ and use the property of logarithms to move that $\dfrac{1}{2}$ to the front. After that, integrate: compute
$$\begin{array}{ll}
\dfrac{\mathrm{d}}{\mathrm{d}x} \ln \left( \sqrt{ \dfrac{x+1}{x-1} } \right) &= \dfrac{\mathrm{d}}{\mathrm{d}x} \ln \left( \left( \dfrac{x+1}{x-1} \right)^{\frac{1}{2}} \right) \\
&= \dfrac{1}{2} \dfrac{\mathrm{d}}{\mathrm{d}x} \ln \left( \dfrac{x+1}{x-1} \right) \\
&= \dfrac{1}{2} \dfrac{1}{\frac{x+1}{x-1}} \dfrac{\mathrm{d}}{\mathrm{d}x} \dfrac{x+1}{x-1} \\
&= \dfrac{1}{2} \dfrac{x-1}{x+1} \dfrac{(x-1)-(x+1)}{(x-1)^2} \\
&= \dfrac{1}{2} \dfrac{1}{x+1} \dfrac{-2}{x-1} \\
&= -\dfrac{1}{x^2-1} \\
&= \dfrac{1}{1-x^2}. \blacksquare
\end{array}$$
Section 5.2 #8) Compute $\displaystyle\int \dfrac{x^2}{5-x^3} \mathrm{d}x$.
Solution: Let $u=5-x^3$ so that $\mathrm{d}u=-3x^2 \mathrm{d}x$. Express that as $-\dfrac{1}{3} \mathrm{d}u=x^2\mathrm{d}x$. Now compute
$$\begin{array}{ll}
\displaystyle\int \dfrac{x^2}{5-x^3} \mathrm{d}x &= -\dfrac{1}{3} \displaystyle\int \dfrac{1}{u} \mathrm{d}u \\
&=-\dfrac{1}{3} \log(u) + C \\
&= -\dfrac{1}{3} \log(5-x^3) + C. \blacksquare
\end{array}$$
Section 5.2 #31) Compute $\displaystyle\int \cot \left( \dfrac{\theta}{3} \right) \mathrm{d}\theta$.
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
$$\begin{array}{ll}
\displaystyle\int \cot \left( \dfrac{\theta}{3} \right) \mathrm{d}\theta &= \displaystyle\int \dfrac{\cos(\frac{\theta}{3})}{\sin(\frac{\theta}{3})} \mathrm{d}\theta \\
&= 3 \displaystyle\int \dfrac{1}{u} \mathrm{d}u \\
&= 3 \ln (|u|) + C \\
&= 3 \ln \left( \left| \sin \left( \dfrac{\theta}{3} \right) \right| \right) + C. \blacksquare
\end{array}$$
Section 5.2 #49) Compute $\displaystyle\int_0^4 \dfrac{5}{3x+1} \mathrm{d}x$.
Solution:
Section 5.2 #55) Compute $\displaystyle\int_1^2 \dfrac{1-\cos(\theta)}{\theta-\sin(\theta)} \mathrm{d}\theta$.
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
$$\begin{array}{ll}
\displaystyle\int_1^2 \dfrac{1-\cos(\theta)}{\theta-\sin(\theta)} \mathrm{d}\theta &= \displaystyle\int_{1-\sin(1)}^{2-\sin(2)} \dfrac{1}{u} \mathrm{d}u \\
&= \ln (|u|) \Bigg|_{1-\sin(1)}^{2-\sin(2)} \\
&= \ln(|2-\sin(2)|) - \ln(|1-\sin(1)|). \blacksquare
\end{array}$$
