$$\begin{gathered} {\int }_0^1{\cot ^-1(x^2+x+1)}^1\text{dx is equal to} \\ \end{gathered}$$

The correct answer is:

{\int }_0^1{\tan ^-1(x+1)}^1\text{dx −}{\int }_0^1{\tan ^-1x}^1\text{dx}

Step-by-step Explanation:

Step 1: Convert cot⁻¹ to tan⁻¹
Recall the inverse trigonometric identity:

\cot ^-1\left(\theta \right)=\tan ^-1\left(\frac{1}{\theta }\right)

for any positive value of \theta.
Since x^2+x+1>0 for all x\in [0,1], we can rewrite the integrand as:

\cot ^-1(x^2+x+1)=\tan ^-1\left(\frac{1}{x^2+x+1}\right)

Step 2: Manipulate the algebraic expression inside tan⁻¹
We can rewrite the denominator x^2+x+1 as 1+x(x+1).
Notice that the numerator 1 can be expressed as the difference (x+1)-x.
Therefore, the argument becomes:

\frac{1}{x^2+x+1}=\frac{(x+1)-x}{1+(x+1)x}

Step 3: Apply the subtraction formula for inverse tangent
Using the standard identity:

\tan ^-1\left(\frac{a-b}{1+ab}\right)=\tan ^-1\left(a\right)-\tan ^-1\left(b\right)

Letting a=x+1 and b=x, we obtain:

\tan ^-1\left(\frac{(x+1)-x}{1+(x+1)x}\right)=\tan ^-1(x+1)-\tan ^-1\left(x\right)

Step 4: Substitute back into the definite integral
Replacing the integrand in the given integral, we get:

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

By the linearity of integrals, this splits into:

{\int }_0^1\tan ^-1(x+1)\text{dx}-{\int }_0^1\tan ^-1\left(x\right)\text{dx}

This matches the first option.