# Prove the following$\int\limits_1^3\frac{dx}{x^2(x+1)}=\frac{2}{3}+log\frac{2}{3}$

Toolbox:
• If the given integral function in a rational function of the form $\large\frac{A}{(x+1)(x+2)^2}=\frac{A}{(x+_1)}+\frac{B}{(x+2)}+\frac{c}{(x+2)^2}$
• (ii)$\int \frac{dx}{(x+a)} =\log (x+a) +c$
• (iii)$\int \limits_a^b f(x)dx=F(b)-F(a)$
$\int \limits_1^3 \frac{dx}{x^2(x+1)}=\frac{2}{3}+log \frac{2}{3}$

given $I=\int \limits_1^3 \frac{dx}{x^2(x+1)}$

This is a proper rational function,hence it can be resolved into its partial function as

$\large\frac{1}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}$

$1=Ax(x+1)+B(x+1)+B+Cx^2$

$=Ax^2+A+Bx+B+Cx^2$

Equating the coefficient of $x^2,$

$0=A+C$ ----------(1)

Equating the coefficient of $x$

$0=A+B$----------(2)

Equating the constant term,

$1=B \qquad Hence\; B=1$

Substituting for B in equ (2) we get,

$A+1=0 => A =-1$

Substituting for A in equ (1)

$A+C=0$

$-1+C=0 => C=1$

Hence $A =-1,B=1,C=1$

Hence on substituting for A,B and C we get

$\frac{1}{x^2(x+1)}=-\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x+1}$

Therefore $I=\int \limits_1^3 \frac{-1}{x} dx+\int \limits_1^3 \frac{1}{x^2}dx+\int \limits_1^3 \frac{a}{x+1} dx$

On integrating we get

$I=\bigg[-log x\bigg]_1^3+\bigg[\frac{x^{-2+1}}{-2+1}\bigg]_1^3+\bigg[log (x+1)\bigg]_1^3dx$

$=[-log x]_1^3-[\frac{1}{x}]_1^3+[log (x+1)]_1^3$

On applying limits,

$-(log 3- log 1)-(\frac{1}{3}-1)+[log[(3+1)-log 2]]$

But $log (a/b) =log a-log b$

Therefore $-(log 3/1)+\frac{2}{3}+log \frac{4}{2}$

$=\frac{2}{3}+\log (\frac{4}{2} /\frac{3}{1})$

$=\frac{2}{3}+log (\frac{2}{3})$

Hence $I=\frac{2}{3}+log (\frac{2}{3})$

Hence proved

answered Mar 7, 2013 by