<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand\ds{\displaystyle}
\begin{document}

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QUESTION

\begin{description}

\item[(a)]
Find the following indefinite integrals

${\rm (i)}\ \ds\int\frac{1}{x+3}\,dx,\hspace{.4cm}{\rm (ii)}\
\int\frac{x+1}{x^2+2x+2}\,dx,\hspace{.4cm}{\rm (iii)}\
\int\frac{1}{x^2+2x+2}\,dx.$

\item[(b)]
Using partial fractions and the results in part (a) of this
question, if appropriate, find

$$\int\frac{x^2-2x-5}{(x+3)(x^2+2x+2)}\,dx.$$

\end{description}

\bigskip

ANSWER
\begin{description}

\item[(a)]

\begin{description}

\item[(i)]
$$\int\frac{dx}{x+3}=\ln|x+3|+c$$

\item[(ii)]
\begin{eqnarray*}
\int\frac{x+1}{x^2+2x+2}\,dx
&amp;=&amp;\int\frac{\frac{1}{2}(2x+2)}{x^2+2x+2}\,dx\\
&amp;=&amp;\frac{1}{2}\ln|x^2+2x+2|+c
\end{eqnarray*}

\item[(iii)]
\begin{eqnarray*}
\int\frac{dx}{x^2+2x+2} &amp;=&amp;\int\frac{dx}{(x+1)^2+1^2}\\
&amp;=&amp;\tan^{-1}(x+1)+c
\end{eqnarray*}

\end{description}

\item[(b)]

\begin{eqnarray*}
\frac{x^2-2x-5}{(x+3)(x^2+2x+2)}
&amp;=&amp;\frac{A}{x+3}+\frac{Bx+C}{x^2+2x+2}\\
&amp;=&amp;\frac{A(x^2+2x+2)+(Bx+C)(x+3)}{(x+3)(x^2+2x+2)}
\end{eqnarray*}

$x^2-2x-5=A(x^2+2x+2)+(Bx+C)(x+3)\\ x=-3;\ 9+6-5=A(9-6+2)+0,\
10=5A,\ A=2\\ x^2;\ 1=A+B,\ B=1-A=-1\\ \textrm{const.};\
-5=2A+3C,\ 3C=-5-2A=-9,\ C=-3$

Thus

\begin{eqnarray*}
&amp;&amp;\int\frac{x^2-2x-5}{(x+3)(x^2+2x+2)}\,dx\\
&amp;&amp;=\int\left(\frac{2}{x+3}-\frac{x+3}{x^2+2x+2}\right)\,dx\\
&amp;&amp;=\int\frac{2}{x+3}\,dx-\int\frac{x+1}{x^2+2x+2}\,dx-\int\frac{2}{x^2+2x+2}\,dx\\
&amp;&amp;=2\ln|x+3|-\frac{1}{2}\ln|x^2+2x+2|-2\tan^{-1}(x+1)+c
\end{eqnarray*}

\end{description}





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