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

1 Answer

  • 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
Equating the coefficient of $x^2,$
$0=A+C$ ----------(1)
Equating the coefficient of $x$
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)
$-1+C=0 => C=1$
Hence $A =-1,B=1,C=1$
Hence on substituting for A,B and C we get
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 meena.p