_2^3 x (x+2)(x+3) d x - Vidyadip

Question

$\int_2^3 \frac{x}{(x+2)(x+3)} d x$

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Answer

Let $I =\int_2^3 \frac{x}{(x+2)(x+3)} \cdot d x$
Let $\frac{x}{(x+2) x+3}=\frac{ A }{x+2}+\frac{ B }{x+3}$
$
\therefore x = A ( x +3)+ B ( x +2)
$
Putting $x=-3$ in (ii) we get
$
\begin{aligned}
& -2= A \\
& \therefore B =3
\end{aligned}
$
Putting $x=-2$ in (ii), we get
$
\begin{aligned}
& -2= A \\
& \therefore A =-2
\end{aligned}
$
From (i), we get
$
\begin{aligned}
& \frac{x}{(x+2(x+3))}=\frac{-2}{x+2}+\frac{3}{x+3} \\
& \therefore I =\int_2^3\left[\frac{-2}{x+2}+\frac{3}{x+3}\right] \cdot d x \\
& =-2 \int_2^3 \frac{1}{x+2} \cdot d x+3 \int_2^3 \frac{1}{x+3} \cdot d x \\
& =-2[\log |x+2|]_2^3+3[\log |x+3|]_2^3 \\
& =-2 \log [\log 5-\log 4]+3[\log 6-\log 5] \\
& =-2\left[\log \left(\frac{5}{4}\right)\right]+3\left[\log \left(\frac{6}{5}\right)\right] \\
& =3 \log \left(\frac{6}{5}\right)-2 \log \left(\frac{5}{4}\right)
\end{aligned}
$
$\begin{aligned} & =\log \left(\frac{6}{5}\right)^2-2 \log \left(\frac{5}{4}\right)^2 \\ & =\log \left(\frac{216}{125}\right)-\log \left(\frac{25}{16}\right) \\ & =\log \left(\frac{216}{125} \times \frac{16}{25}\right) \\ & \therefore I =\log \left(\frac{3456}{3125}\right) .\end{aligned}$

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