Key Concepts and Formulas

• Inverse Trigonometric Identity: $\cot^{-1}y = \frac{\pi}{2} - \tan^{-1}y$ for any real number $y$.
• Integration by Parts: $\int u \,dv = uv - \int v \,du$.
• Substitution Method: Used for integrals of the form $\int f(g(x)) g'(x) \,dx$.
• Definite Integral Properties: Linearity of integration and evaluation at limits.

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Step-by-Step Solution

We are given the equation: $2\int\limits_0^1 {{{\tan }^{ - 1}}xdx = \int\limits_0^1 {{{\cot }^{ - 1}}} } \left( {1 - x + {x^2}} \right)dx$ We need to find the value of $\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx$.

Step 1: Utilize the Inverse Trigonometric Identity

The presence of $\cot^{-1}(1-x+x^2)$ in the given equation and $\tan^{-1}(1-x+x^2)$ in the integral we need to find suggests using the identity $\cot^{-1}y = \frac{\pi}{2} - \tan^{-1}y$. Let $y = 1 - x + x^2$. Applying this identity to the RHS of the given equation: $\int\limits_0^1 {{{\cot }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx = \int\limits_0^1 \left( {\frac{\pi }{2} - {{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)} \right)dx$ Using the linearity of integration, we can split this into two integrals: $= \int\limits_0^1 {\frac{\pi }{2}} dx - \int\limits_0^1 {{{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)} dx$ Now, substitute this back into the original given equation: $2\int\limits_0^1 {{{\tan }^{ - 1}}xdx = \int\limits_0^1 {\frac{\pi }{2}} dx - \int\limits_0^1 {{{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)} dx}$

Step 2: Isolate the Target Integral

Our goal is to find $\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx$. We can rearrange the equation from Step 1 to group this integral. Move the term $\int\limits_0^1 {{{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)} dx$ to the left-hand side and $2\int\limits_0^1 {{{\tan }^{ - 1}}xdx}$ to the right-hand side: $\int\limits_0^1 {{{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)} dx = \int\limits_0^1 {\frac{\pi }{2}} dx - 2\int\limits_0^1 {{{\tan }^{ - 1}}x} dx$ First, evaluate the integral of a constant: $\int\limits_0^1 {\frac{\pi }{2}} dx = \left[ {\frac{\pi }{2}x} \right]_0^1 = \frac{\pi}{2}(1) - \frac{\pi}{2}(0) = \frac{\pi}{2}$ So, the equation becomes: $\int\limits_0^1 {{{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)} dx = \frac{\pi}{2} - 2\int\limits_0^1 {{{\tan }^{ - 1}}x} dx \quad (*)$ To proceed, we need to evaluate the integral $\int\limits_0^1 {{{\tan }^{ - 1}}x} dx$.

Step 3: Evaluate $\int\limits_0^1 {{{\tan }^{ - 1}}x} dx$ using Integration by Parts

Let $I = \int\limits_0^1 {{{\tan }^{ - 1}}x} dx$. We use integration by parts with $u = \tan^{-1}x$ and $dv = dx$. Then, $du = \frac{1}{1+x^2} dx$ and $v = x$. Applying the integration by parts formula: $I = \left[ {x{{\tan }^{ - 1}}x} \right]_0^1 - \int\limits_0^1 {x \cdot \frac{1}{1+x^2}} dx$ Evaluate the first term: $\left[ {x{{\tan }^{ - 1}}x} \right]_0^1 = (1 \cdot \tan^{-1}1) - (0 \cdot \tan^{-1}0) = 1 \cdot \frac{\pi}{4} - 0 = \frac{\pi}{4}$ Now, evaluate the remaining integral: $\int\limits_0^1 {\frac{x}{1+x^2}} dx$. We use the substitution method. Let $t = 1+x^2$. Then $dt = 2x \,dx$, which implies $x \,dx = \frac{1}{2}dt$. Change the limits of integration: When $x=0$, $t = 1+0^2 = 1$. When $x=1$, $t = 1+1^2 = 2$. The integral becomes: $\int\limits_1^2 {\frac{1}{t}} \cdot \frac{1}{2} dt = \frac{1}{2} \int\limits_1^2 {\frac{1}{t}} dt = \frac{1}{2} \left[ {\log |t|} \right]_1^2$ $= \frac{1}{2} (\log 2 - \log 1) = \frac{1}{2} (\log 2 - 0) = \frac{1}{2}\log 2$ Substituting these results back into the expression for $I$: $I = \frac{\pi}{4} - \frac{1}{2}\log 2$

Step 4: Substitute and Find the Final Answer

Now, substitute the value of $\int\limits_0^1 {{{\tan }^{ - 1}}x} dx$ back into equation $(*)$: $\int\limits_0^1 {{{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)} dx = \frac{\pi}{2} - 2 \left( {\frac{\pi}{4} - \frac{1}{2}\log 2} \right)$ Distribute the $2$: $= \frac{\pi}{2} - \left( {2 \cdot \frac{\pi}{4} - 2 \cdot \frac{1}{2}\log 2} \right)$ $= \frac{\pi}{2} - \left( {\frac{\pi}{2} - \log 2} \right)$ $= \frac{\pi}{2} - \frac{\pi}{2} + \log 2$ $= \log 2$

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Common Mistakes & Tips

• Incorrectly applying the $\cot^{-1}$ identity: Ensure the identity $\cot^{-1}y = \frac{\pi}{2} - \tan^{-1}y$ is used correctly, especially regarding the sign.
• Errors in Integration by Parts: For definite integrals, remember to evaluate the $uv$ term at both limits. Also, choose $u$ and $dv$ strategically for inverse trigonometric functions.
• Substitution Errors: When using substitution in a definite integral, remember to change the limits of integration according to the substitution, or substitute back to the original variable before evaluating.

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Summary

The problem was solved by first using the identity $\cot^{-1}y = \frac{\pi}{2} - \tan^{-1}y$ to transform the given equation. This allowed us to express the integral we needed to find in terms of the integral of $\tan^{-1}x$. The integral of $\tan^{-1}x$ was then evaluated using integration by parts. Finally, substituting the result back yielded the required value.

The final answer is $\boxed{\text{log2}}$, which corresponds to option (C).