Integrals — Class 12 Maths Solution

Option b is correct

Let $I = \int\limits_0^1 {{{\tan }^{ - 1}}\left( {\cfrac{{2x - 1}}{{1 + x - {x^2}}}} \right)dx}$

$= \int\limits_0^1 {{{\tan }^{ - 1}}\left[ {\cfrac{{x + x - 1}}{{1 - x\left( {x - 1} \right)}}} \right]dx} = \int\limits_0^1 {\left[ {{{\tan }^{ - 1}}x + {{\tan }^{ - 1}}\left( {x - 1} \right)} \right]dx}$

$\Rightarrow$ $I = \int\limits_0^1 {{{\tan }^{ - 1}}xdx} + \int\limits_0^1 {{{\tan }^{ - 1}}\left( {x - 1} \right)dx} = {I_1} + {I_2}$

….(i)
Where, ${I_1} = \int\limits_0^1 {1 \cdot {{\tan }^{ - 1}}x} dx$

Integrating by parts
$= \left[ {{{\tan }^{ - 1}}x\int {\left( 1 \right)dx} - \int {\left( {\cfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) \cdot \int {\left( 1 \right)} dx} \right)} dx} \right]_0^1$

$= \left[ {x{{\tan }^{ - 1}}x - \int {\cfrac{1}{{1 + {x^2}}} \times x} dx} \right]_0^1$

$= \left[ {x{{\tan }^{ - 1}}x - \cfrac{1}{2}\int {\cfrac{{2x}}{{1 + {x^2}}}} dx} \right]_0^1$

$= \left[ {x{{\tan }^{ - 1}}x - \cfrac{1}{2}\log \left( {1 + {x^2}} \right)} \right]_0^1$

$= \left[ {\left( {{{\tan }^{ - 1}}1 - 0} \right) - \cfrac{1}{2}\left( {\log 2 - \log 1} \right)} \right]$
$= \left[ {\cfrac{\pi }{4} - \cfrac{1}{2}\log 2} \right]$

…..(ii)
And ${I_2} = \int\limits_2^1 {1 \cdot {{\tan }^{ - 1}}\left( {x - 1} \right)dx}$

Again integrating by parts
$= \left[ {{{\tan }^{ - 1}}\left( {x - 1} \right)\int {\left( 1 \right)dx - \int {\left( {\cfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}\left( {x - 1} \right) \cdot \int {\left( 1 \right)dx} } \right)dx} \right)} } } \right]_0^1$

$= \left[ {{{\tan }^{ - 1}}\left( {x - 1} \right) \cdot x - \int {\left( {\cfrac{1}{{1 + {{\left( {x - 1} \right)}^2}}} \times x} \right)dx} } \right]_0^1$

$= \left[ {x{{\tan }^{ - 1}}\left( {x - 1} \right) - \cfrac{1}{2}\int {\cfrac{{2\left( {x - 1 + 1} \right)}}{{1 + {{\left( {x - 1} \right)}^2}}}dx} } \right]_0^1$

$= \left[ {x{{\tan }^{ - 1}}\left( {x - 1} \right) - \cfrac{1}{2}\int {\cfrac{{2\left( {x - 1} \right)dx}}{{1 + {{\left( {x - 1} \right)}^2}}} - \cfrac{1}{2}\int {\cfrac{2}{{1 + {{\left( {x - 1} \right)}^2}}}dx} } } \right]_0^1$

$= \left[ {x{{\tan }^{ - 1}}\left( {x - 1} \right)} \right]_0^1 - \cfrac{1}{2}\left[ {\log \left( {1 + {{\left( {x - 1} \right)}^2}} \right)} \right]_0^1 - \left[ {{{\tan }^{ - 1}}\left( {x - 1} \right)} \right]_0^1$

$= \left[ {0 - 0} \right] - \cfrac{1}{2}\left[ {0 - \log 2} \right] - \left[ {0 + \cfrac{\pi }{4}} \right] = \cfrac{1}{2}\log 2 - \cfrac{\pi }{4}$

…(iii)
From (i), (ii) and (iii)

we get
$I = \left[ {\cfrac{\pi }{4} - \cfrac{1}{2}\log 2 + \cfrac{1}{2}\log 2 - \cfrac{\pi }{4}} \right] = 0$