Evaluate \[\int\limits_0^1 {{{\cot }^{ - 1}}(1 - x + {x^2})dx} \]
A.\[\pi - \dfrac{{\log }}{2}\]
B. \[2\pi - \dfrac{{\log }}{2}\]
C. \[\pi + \dfrac{{\log }}{2}\]
D. None of these

Hint: We use the property that tangent and cotangent functions are reciprocal of each other. Convert the integral in terms of inverse tangent function and break the numerator in such a way that we can apply the property of addition of two tangent inverse functions. Use integration by parts to find the value of the obtained integral.
* \[\because \cot \theta = \dfrac{1}{{\tan \theta }} \Rightarrow {\cot ^{ - 1}}\theta = {\tan ^{ - 1}}\dfrac{1}{\theta }\]
* \[{\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)\]
*In integration by parts we use the sequence of ILATE (Inverse trigonometric, Logarithms, algebraic, trigonometric, and exponential) for choosing the value of \[u,v\] and then substitute in the integration formula of by parts integration.

Complete step by step solution:
We have to evaluate the integral \[\int\limits_0^1 {{{\cot }^{ - 1}}(1 - x + {x^2})dx} \]
Convert the integral in terms of tangent inverse
\[ \Rightarrow {\cot ^{ - 1}}(1 - x + {x^2})dx = \int\limits_0^1 {{{\tan }^{ - 1}}\dfrac{1}{{(1 - x + {x^2})}}} dx\]................… (1)
Now we break the numerator in such a way so we can apply property of trigonometry
We can write \[1 = x + 1 - x\]
Substitute the value of \[1 = x + 1 - x\] in numerator of equation (1)
\[ \Rightarrow {\cot ^{ - 1}}(1 - x + {x^2})dx = \int\limits_0^1 {{{\tan }^{ - 1}}\dfrac{{x + 1 - x}}{{(1 - x + {x^2})}}} dx\]
Write the denominator in easy terms
\[ \Rightarrow {\cot ^{ - 1}}(1 - x + {x^2})dx = \int\limits_0^1 {{{\tan }^{ - 1}}\dfrac{{x + 1 - x}}{{1 - x(1 - x)}}} dx\]...............… (2)
If we take two angles, \[(x)\] and \[(1 - x)\] then we can write \[{\tan ^{ - 1}}(x) + {\tan ^{ - 1}}(1 - x) = {\tan ^{ - 1}}\left( {\dfrac{{x + (1 - x)}}{{1 - x(1 - x)}}} \right)\]
So, RHS of the equation (2) transforms into easier form using the identity \[{\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)\]
\[ \Rightarrow {\cot ^{ - 1}}(1 - x + {x^2})dx = \int\limits_0^1 {\left( {{{\tan }^{ - 1}}(x) + {{\tan }^{ - 1}}(1 - x)} \right)dx} \]
Separate the two inverse functions in RHS
\[ \Rightarrow {\cot ^{ - 1}}(1 - x + {x^2})dx = \int\limits_0^1 {{{\tan }^{ - 1}}(x)dx} + \int\limits_0^1 {{{\tan }^{ - 1}}(1 - x)dx} \].................… (3)
Since we know that \[\int\limits_0^a {f(x)} dx = \int\limits_0^a {f(a - x)} dx\]
So, \[\int\limits_0^1 {{{\tan }^{ - 1}}(1 - x)dx} = \int\limits_0^1 {{{\tan }^{ - 1}}1 - (1 - x)dx} \]
Which on solving gives \[\int\limits_0^1 {{{\tan }^{ - 1}}(1 - x)dx} = \int\limits_0^1 {{{\tan }^{ - 1}}\left( {1 - 1 + x} \right)dx} \]
i.e. \[\int\limits_0^1 {{{\tan }^{ - 1}}(1 - x)dx} = \int\limits_0^1 {{{\tan }^{ - 1}}xdx} \]
Substitute the value of \[\int\limits_0^1 {{{\tan }^{ - 1}}(1 - x)dx} = \int\limits_0^1 {{{\tan }^{ - 1}}xdx} \] in equation (3)
\[ \Rightarrow {\cot ^{ - 1}}(1 - x + {x^2})dx = \int\limits_0^1 {{{\tan }^{ - 1}}(x)dx} + \int\limits_0^1 {{{\tan }^{ - 1}}(x)dx} \]
Add like terms in RHS
\[ \Rightarrow {\cot ^{ - 1}}(1 - x + {x^2})dx = 2\int\limits_0^1 {{{\tan }^{ - 1}}(x)dx} \]
Now we can write RHS as
\[ \Rightarrow {\cot ^{ - 1}}(1 - x + {x^2})dx = 2\int\limits_0^1 {{{\tan }^{ - 1}}x \times 1dx} \]
Use by parts to integrate RHS
Looking at the integral \[2\int\limits_0^1 {{{\tan }^{ - 1}}x \times 1dx} \]
Using the ILATE sequence we can tell that the term comes first in the sequence is \[u\] and which comes later is \[v\] i.e. in this case we have the terms inverse trigonometric and algebraic , first comes the inverse trigonometric term then comes the algebraic term so \[u\] is \[{\tan ^{ - 1}}x\] and \[v\] is 1
\[u = {\tan ^{ - 1}}x,v = 1\]
Now we calculate the differentiation of the term \[u\]
\[u' = \dfrac{{du}}{{dx}}\]
\[ \Rightarrow u' = \dfrac{{d({{\tan }^{ - 1}}x)}}{{dx}}\]
\[ \Rightarrow u' = \dfrac{1}{{1 + {x^2}}}\]
Now substitute the values in equation \[\int {uvdt = u\int {vdt - \int {u'\left( {\int {vdt} } \right)dt} } } \]
\[ \Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2{\tan ^{ - 1}}x\int\limits_0^1 {1dx} - 2\int\limits_0^1 {(\dfrac{1}{{1 + {x^2}}})\left( {\int {1dx} } \right)dx} \]
Now we know \[\int {1dx = x} \]
Therefore substitute the value of \[\int {1dx = x} \] in RHS
\[ \Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2{\tan ^{ - 1}}x(x) - 2\int\limits_0^1 {(\dfrac{1}{{1 + {x^2}}})(x)dx} \]
Now we can solve and bring out constant terms from the integral sign.
\[ \Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2x{\tan ^{ - 1}}x - \int\limits_0^1 {(\dfrac{{2x}}{{1 + {x^2}}})dx} \]
Since the numerator of the fraction in RHS is derivative of the denominator and we know \[\int {\dfrac{{dt}}{t} = \log t} \]
\[ \Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2\left[ {x{{\tan }^{ - 1}}x} \right]_0^1 - \left[ {\log \left( {1 + {x^2}} \right)} \right]_0^1\]
Calculate the values by substituting the upper and lower limits
\[ \Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2\left[ {1{{\tan }^{ - 1}}1 - 0} \right] - \left[ {\log \left( {1 + 1} \right) - \log \left( {1 + 0} \right)} \right]\]
\[ \Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2\left[ {{{\tan }^{ - 1}}1} \right] - \left[ {\log 2 - \log 1} \right]\]
Substitute \[{\tan ^{ - 1}}1 = \dfrac{\pi }{4}\] and use the value \[\log 1 = 0\]
\[ \Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2 \times \dfrac{\pi }{4} - \left[ {\log 2 - 0} \right]\]
\[ \Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = \dfrac{\pi }{2} - \log 2\]
So value of \[2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} \] is \[\dfrac{\pi }{2} - \log 2\].

\[\therefore \]Option D is correct.

Note: Alternative method:
We can solve the solution after the step \[2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2\left[ {{{\tan }^{ - 1}}1} \right] - \left[ {\log 2 - \log 1} \right]\] in another way
For step in the end we can use rule of log i.e. \[\log m - \log n = \log \dfrac{m}{n}\]
\[ \Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = 2\left[ {{{\tan }^{ - 1}}1} \right] - \left[ {\log \dfrac{2}{1}} \right]\]
\[ \Rightarrow 2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} = \dfrac{\pi }{2} - \log 2\]
So value of \[2\int\limits_0^1 {{{\tan }^{ - 1}}x.1dx} \]is\[\dfrac{\pi }{2} - \log 2\].
\[\therefore \]Option D is correct.