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	{\noindent 21-120 Differential and Integral Calculus \hfill Summer Session I 2010\\}
	{\centering Quiz 15\medskip\\}
	\hrule
	{
		\begin{prob*} \mbox{}
			\begin{description}
				\item[1.] $\ds\int\frac{x}{x-6}\;dx$
				\item[Bonus.] $\ds\int\arctan\left(\frac{1}{x}\right)dx$
			\end{description}
		\end{prob*}
		\begin{soln*} \mbox{}
			\begin{description}
				\item[1.] We do the substitution $u=x-6$. Then $du=dx$ and $x=u+6$. Thus we have:
				\begin{align*}
					\int\frac{x}{x-6}\;dx 	&= \int\frac{u+6}{u} du \\
									&= \int\left(\frac{u}{u} + \frac{6}{u}\right)du\\
									&= \int\left(1 + \frac{6}{u}\right)du\\
									&= u + \ln|u| + C\\
									&= x-6 + \ln|x-6|+C\\
									&= x + \ln|x-6| + C
				\end{align*}
				\item[Bonus] We begin with parts:
				\begin{align*}
				\int\arctan\left(\frac{1}{x}\right)dx \\
				\begin{array}[h]{l l}
					u=\arctan\left(\frac{1}{x}\right) & dv = dx \\
					du= \frac{1}{1+(1/x)^2}\cdot \left(-\frac{1}{x^2}\right) & v = x
				\end{array} \\
				&= \arctan\left(\frac{1}{x}\right)\cdot x - \int \left(\frac{1}{1+(1/x)^2} \cdot \left(-\frac{x}{x^2}\right) \right) dx\\
				&= x\arctan\left(\frac{1}{x}\right) + \int \left(\frac{x}{x^2+1} \right) dx\\
				\begin{array}[h]{r l}
				  	u =& x^2 + 1\\
				  	du =& 2x\cdot dx \\
				  	\frac{du}{2} =& x\cdot dx
				  \end{array} \\
				  &= x\arctan\left(\frac{1}{x}\right) + \frac{1}{2}\int \left(\frac{1}{u} \right) du\\
				  &= x\arctan\left(\frac{1}{x}\right) + \frac{\ln|x^2+1|}{2} + C\\
				\end{align*}

			\end{description}

		\end{soln*}

	}
\end{document}