_ ^ x^2 + 1 [ ( x^2 + 1) - 2 x] x^4 dx is equal to - Vidyadip

MCQ

$\int_{}^{} {\frac{{\sqrt {{x^2} + 1} [\log ({x^2} + 1) - 2\log x]}}{{{x^4}}}} dx$ is equal to

A
$\frac{1}{3}{\left( {1 + \frac{1}{{{x^2}}}} \right)^{1/2}}\left[ {\log \left( {1 + \frac{1}{{{x^2}}}} \right) + \frac{2}{3}} \right] + c$
✓
$ - \frac{1}{3}{\left( {1 + \frac{1}{{{x^2}}}} \right)^{3/2}}\left[ {\log \left( {1 + \frac{1}{{{x^2}}}} \right) - \frac{2}{3}} \right] + c$
C
$\frac{2}{3}{\left( {1 + \frac{1}{{{x^2}}}} \right)^{3/2}}\left[ {\log \left( {1 + \frac{1}{{{x^2}}}} \right) + \frac{2}{3}} \right] + c$
D
None of these

✓

Answer

Correct option: B.

$ - \frac{1}{3}{\left( {1 + \frac{1}{{{x^2}}}} \right)^{3/2}}\left[ {\log \left( {1 + \frac{1}{{{x^2}}}} \right) - \frac{2}{3}} \right] + c$

b
(b) $\int_{}^{} {\frac{{\sqrt {{x^2} + 1} \left[ {\log ({x^2} + 1) - 2\log x} \right]}}{{{x^4}}}\,dx} $
$ = \int_{}^{} {\sqrt {1 + \frac{1}{{{x^2}}}} \log \left( {1 + \frac{1}{{{x^2}}}} \right)\,.\,\frac{1}{{{x^3}}}\,dx} $
$\left\{ {{\rm{Put}}\,1 + \frac{1}{{{x^2}}} = t \Rightarrow - \frac{2}{{{x^3}}}\,dx = dt} \right\}$
$ = - \frac{1}{2}\int_{}^{} {\sqrt t \log t\,dt} = - \frac{1}{2}\left\{ {\log t\left( {\frac{{{t^{3/2}}}}{{3/2}}} \right) - \frac{2}{3}\int_{}^{} {\frac{1}{t}{t^{3/2}}dt} } \right\}$
$ = - \frac{1}{2}\left\{ {\frac{2}{3}\log t\,.\,{t^{3/2}} - \frac{2}{3}.\,\left( {\frac{2}{3}{t^{3/2}}} \right)} \right\} + c$
$ = - \frac{1}{3}{t^{3/2}}\log t + \frac{2}{9}{t^{3/2}} + c$
$ = - \frac{1}{3}{\left( {1 + \frac{1}{{{x^2}}}} \right)^{3/2}}\left[ {\log \left( {1 + \frac{1}{{{x^2}}}} \right) - \frac{2}{3}} \right] + c$.

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