The value of _2^3 x + 1 x^2 (x - 1) dx is - Vidyadip

MCQ

The value of $\int_2^3 {\frac{{x + 1}}{{{x^2}(x - 1)}}dx} $ is

A
$2\log 2 - \frac{1}{6}$
✓
$\log \frac{{16}}{9} - \frac{1}{6}$
C
$\log \frac{4}{3} - \frac{1}{6}$
D
$\log \frac{{16}}{9} + \frac{1}{6}$

✓

Answer

Correct option: B.

$\log \frac{{16}}{9} - \frac{1}{6}$

b
(b) $I = \int_2^3 {\frac{{x + 1}}{{{x^2}(x - 1)}}} dx = \int_2^3 {\left( {\frac{A}{{{x^2}}} + \frac{B}{x} + \frac{C}{{x - 1}}} \right)} \,dx$

$A(x - 1) + B(x)(x - 1) + C({x^2}) = x + 1$

Put $x = 0,\,1,\, - 1,$

we get $A = - 1,B = - 2,C = 2$

==> $I = - \int_2^3 {\frac{{dx}}{{{x^2}}} - 2\int_2^3 {\frac{{dx}}{x} + 2\int_2^3 {\frac{{dx}}{{x - 1}}} } } $

==> $I = \left[ {\frac{1}{x}} \right]_2^3 - 2[\log x]_2^3 + 2[\log (x - 1)]_2^3$

==> $I = \frac{1}{3} - \frac{1}{2} - 2\log \frac{3}{2} + 2\log 2$

==> $I = \log \frac{{16}}{9} - \frac{1}{6}$.

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