Question

${\int }_2^3\frac{x}{(x+2)(x+3)}dx$

Answer 1

We have,

$I={\int }_2^3\frac{x}{(x+2)(x+3)}dx$

From partial fraction,

$\frac{x}{(x+2)(x+3)}=\frac{A}{(x+2)}+\frac{B}{(x+3)}$

Then, Solving and we get,

$A=-2andB=3$

Therefore,

$I={\int }_2^3\frac{x}{(x+2)(x+3)}dx={\int }_2^3\frac{-2}{(x+2)}dx+{\int }_2^3\frac{3}{(x+3)}dx$

On integrating and we get,

$I={\int }_2^3\frac{x}{(x+2)(x+3)}dx=-2{{\left[\log (x+2)\right]}_2}^3+3{{\left[\log (x+3)\right]}_2}^3$

$I={\int }_2^3\frac{x}{(x+2)(x+3)}dx=-2[\log (3+2)-\log (2+2)]+3[\log (3+3)-\log (2+3)]$

$I={\int }_2^3\frac{x}{(x+2)(x+3)}dx=-2[\log 5-\log 4]+3[\log 6-\log 5]$

$I={\int }_2^3\frac{x}{(x+2)(x+3)}dx=-2\left[\log \frac{5}{4}\right]+3\left[\log \frac{6}{5}\right]$

Hence, this is the answer.