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QUESTION
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\item[(a)]
Using partial fractions evaluate
$\ds\int_1^2\frac{1}{x(x+2)}\,dx.$

\item[(b)]

\begin{description}

\item[(i)]
Show that

$$\frac{d}{dx}(\tanh x)=\frac{1}{\cosh^2x}.$$

\item[(ii)]
Given that $x=2$ is an approximate solution of the equation
$2\tanh x=x$ use the Newton Raphson formula THREE times, and the
result in part (i), to obtain a better approximation (correct to
four decimal places).

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ANSWER

\begin{description}

\item[(a)]
$\ds\int_1^2\frac{1}{x(x+2)}\,dx$\\
$\ds\frac{1}{x(x+2)}=\frac{A}{x}+\frac{B}{x+2}=\frac{A(x+2)+Bx}{x(x+2)}$
therefore $\ds1=A(x+2)+Bx.\\ x=0;\ 1=2A+0,\ A=\frac{1}{2}\\x=-2;\
1=0-2B,\ B=-\frac{1}{2}$\\
 therefore
 \begin{eqnarray*}
 \int_1^2\frac{1}{x(x+2)}\,dx&=& \int_1^2\left\{\frac{\frac{1}{2}}{x}-\frac{\frac{1}{2}}{x+2}\right\}\,dx\\
 &=&\left[\frac{1}{2}\ln x-\frac{1}{2}\ln (x+2)\right]_1^2\\
 &=&\left(\frac{1}{2}\ln 2-\frac{1}{2}\ln 4\right)-\left(\frac{1}{2}\ln
 1-\frac{1}{2}\ln 3\right)\\
 &=&\frac{1}{2}\ln\left(\frac{2\times3}{4}\right)\\
 &=&\frac{1}{2}\ln\left(\frac{3}{2}\right)
 \end{eqnarray*}


\item[(b)]

\begin{description}

\item[(i)]
Now $\ds u=\tanh x=\frac{\sinh x}{\cosh x},\
\frac{du}{dx}=\frac{\cosh x\cosh x-\sinh x\sinh x}{(\cosh
x)^2}=\frac{1}{\cosh^2x}$


\item[(ii)]
$f(x)=2\tanh x-x,\ f'(x)=2\textrm{sech}^2x-1$\\ given $\ds
x_0=2,\\
x_1=2-\frac{f(2)}{f'(2)}=2-\frac{-0.071945}{-0.85870}=1.91622\\
x_2=x_1-\frac{f(x_1)}{f'(x_1)}=1.91622-\frac{-0.00101}{-0.83401}=1.91501\\
x_3=1.91501-\frac{f(1.191501)}{f'(1.91501)}=1.91501-\frac{-1.63\times10^{-6})}
{(-0.83363)}=1.91501$\\
Third approximation $\Rightarrow x=1.9150$ (correct to 4 decimal
places.)

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