Animated Solution for Mathematics - Definite Integration: Prove that ${\int }_0^1{\tan }^{-1}\left(\frac{1}{1-x+x^2}\right)dx=2{\int }_0^1{\tan }^{-1}xdx$. Hence or otherwise, evaluate the integral ${\int }_0^1{\tan }^{-1}\left(1-x+x^2\right)dx$.

The integral we are tasked to solve is:

This expression appears daunting due to the rational function nested within the inverse trigonometric function. However, in JEE Advanced mathematics, such structures often hide elegant symmetries.

First, we decode the denominator $1-x+x^2$. We can rewrite this as $1+x(x-1)$.

Observing the numerator, which is $1$, we can express it as the difference $x-(x-1)$. This allows us to rewrite the integrand as:

This matches the standard identity for the difference of two arctangents:

By setting $A=x$ and $B=x-1$, the integrand simplifies significantly to ${\tan }^{-1}x-{\tan }^{-1}(x-1)$.

We now split the integral $I$ into two parts: $I_1={\int }_0^1{\tan }^{-1}xdx$ and $I_2={\int }_0^1{\tan }^{-1}(x-1)dx$. To evaluate $I_2$, we apply the King's Property:

Substituting $x$ with $(1-x)$ in $I_2$, the argument becomes $(1-x)-1=-x$. Since ${\tan }^{-1}(-x)=-{\tan }^{-1}x$, we find that $I_2=-{\int }_0^1{\tan }^{-1}xdx=-I_1$.

Consequently, the original integral simplifies to:

Next, we consider the integral $J={\int }_0^1{\tan }^{-1}(1-x+x^2)dx$. We utilize the complementary identity ${\tan }^{-1}z+{\cot }^{-1}z=\frac{\pi }{2}$, where ${\cot }^{-1}z={\tan }^{-1}(1/z)$.

This allows us to express the integrand as:

Integrating both sides from $0$ to $1$, we obtain:

We evaluate $I_1={\int }_0^1{\tan }^{-1}xdx$ using integration by parts, where $u={\tan }^{-1}x$ and $dv=dx$:

Finally, substituting $I_1$ back into our expression for $J$:

The final result is: