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QUESTION

\begin{description}

\item[(a)]

\begin{description}

\item[(i)]
Find the indefinite integral $\ds\int\frac{dx}{4+x^2}$.

\item[(ii)]
Using partial fractions and the result in (i), if appropriate,
find $$\int\frac{3}{(1+x^2)(4+x^2)}\,dx.$$

\end{description}

\item[(b)]
Differentiate with respect to $x$ the functions

(i)$\ds\frac{x}{\sqrt{1+x^2}}$,\hspace{1cm}(ii)$\exp(x^2\sinh x).$

\end{description}



ANSWER

\begin{description}

\item[(a)]

\begin{description}

\item[(i)]
$\ds I=\int\frac{dx}{4+x^2}=\int\frac{dx}{4(1+\frac{x^2}{4})}$ Put
$\ds u=\frac{x}{2},\ \frac{du}{dx}=\frac{1}{2}$ therefore\\ $\ds
I=\int\frac{2du}{4(1+u^2)}=\frac{1}{2}\tan^{-1}u+c
=\frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right)+c$

\item[(ii)]
\begin{eqnarray*}
\frac{3}{(4+x^2)(1+x^2)}
&=&\frac{Ax+B}{4+x^2}+\frac{Cx+D}{1+x^2}\\
&=&\frac{(Ax+B)(1+x^2)+(Cx+D)(4+x^2)}{(4+x^2)(1+x^2)}
\end{eqnarray*}
 i.e. $3=(Ax+B)(1+x^2)+(Cx+D)(4+x^2)$

\begin{tabular}{clc}
Coefficients of:\\ $x^3$&A+C=0&(1)\\ $x$&A+4C=0&(2)\\
$x^2$&B+D=0&(3)\\const.&B+4D=3&(4)
\end{tabular}

$(2)-(1)\Rightarrow3C=0\Rightarrow C=0\Rightarrow A=0\\
(4)-(3)\Rightarrow 3D=3\Rightarrow D=1\Rightarrow B=-D=-1$\\
Therefore
$\ds\frac{3}{(4+x^2)(1+x^2)}=\frac{1}{1+x^2}-\frac{1}{4+x^2}$\\
Hence \begin{eqnarray*} \ds\int\frac{3dx}{(4+x^2)(1+x^2)}
&=&\int\frac{1}{1+x^2}\,dx-\int\frac{1}{4+x^2}\,dx\\
&=&\tan^{-1}x-\frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right)+c
\end{eqnarray*}

\end{description}

\item[(b)]

\begin{description}

\item[(i)]

\begin{eqnarray*}
\frac{d}{dx}\left\{\frac{x}{\sqrt{1+x^2}}\right\}
&=&\frac{\sqrt{1+x^2}.1-x\{\frac{1}{2}(1+x^2)^{-\frac{1}{2}}.2x\}}{1+x^2}\\
&=&\frac{\sqrt{1+x^2}-\frac{x^2}{\sqrt{1+x^2}}}{1+x^2}\\
&=&\frac{(1+x^2)-x^2}{(1+x^2)^\frac{3}{2}}\\
&=&\frac{1}{(1+x^2)^\frac{3}{2}}
\end{eqnarray*}

\item[(ii)]
$$\frac{d}{dx}\{e^{x^2\sinh x}\}=e^{x^2\sinh x}\{x^2\cosh
x+2x\sinh x\}$$

\end{description}

\end{description}



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