Question

Prove: $∫_1^3\frac {dx}{x^2(x+1)}=\frac 23+log\frac 23$

Answer 1

Let I=$$∫_1^3\frac {dx}{x^2(x+1)}=\frac 23+log\frac 23

Also, let $\frac {1}{x^2(x+1)}$ = $\frac Ax+\frac {B}{x^2}+\frac {C}{x+1}$

$⇒I=Ax(x+1)+B(x+1)+C(x^2)$

$⇒I=Ax^2+Ax+Bx+Cx^2$

Equating the coefficients of $x^2,x,$ and constant term, we obtain

$A+C=0$

$A+B=0$

$B=1$

On solving these equations, we obtain

$A=-1,C=1,\ and \ B=1$

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

$⇒I$ = ∫_1^3$$[-\frac 1x+\frac {1}{x^2}+\frac {1}{x+1}]dx

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

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

$I$= $log(\frac 43)-\frac 13-log(\frac 21)+1$

$I$= $log\ 4-log\ 3-log\ 2+\frac 23$

$I$= $log\ 2-log\ 3+\frac 23$

$I$= $log(\frac 23)+\frac 23$

Hence, the given result is proved.