_ ^ 1 (x - 1)( x^2 + 1) dx = - Vidyadip

MCQ

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

✓
$\frac{1}{2}\log (x - 1) - \frac{1}{4}\log ({x^2} + 1) - \frac{1}{2}{\tan ^{ - 1}}x + c$
B
$\frac{1}{2}\log (x - 1) + \frac{1}{4}\log ({x^2} + 1) - \frac{1}{2}{\tan ^{ - 1}}x + c$
C
$\frac{1}{2}\log (x - 1) - \frac{1}{2}\log ({x^2} + 1) - \frac{1}{2}{\tan ^{ - 1}}x + c$
D
None of these

✓

Answer

Correct option: A.

$\frac{1}{2}\log (x - 1) - \frac{1}{4}\log ({x^2} + 1) - \frac{1}{2}{\tan ^{ - 1}}x + c$

a
(a) We have $\frac{1}{{(x - 1)({x^2} + 1)}} = \frac{A}{{(x - 1)}} + \frac{{Bx + C}}{{({x^2} + 1)}}$
$ \Rightarrow 1 = A({x^2} + 1) + (Bx + C)(x - 1)$
If $x = 1,$ then $A = \frac{1}{2}$ .....$(i)$
$A - C = 1 \Rightarrow C = - \frac{1}{2}$ .....$(ii)$
$A + B = 0 \Rightarrow B = - \frac{1}{2}$ .....$(iii)$
Putting these values, we get
$\frac{1}{{(x - 1)({x^2} + 1)}} = \frac{1}{2}.\frac{1}{{(x - 1)}} - \frac{{x + 1}}{{2({x^2} + 1)}}$
Hence $\int_{}^{} {\frac{1}{{(x - 1)({x^2} + 1)}}} dx = \frac{1}{2}\int_{}^{} {\frac{{dx}}{{(x - 1)}} - \frac{1}{2}} \,\int_{}^{} {\frac{{x + 1}}{{{x^2} + 1}}} dx$
$ = \frac{1}{2}\log (x - 1) - \frac{1}{4}\log ({x^2} + 1) - \frac{1}{2}{\tan ^{ - 1}}x + c.$

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