The value of $\int \cot^{-1}\left(\frac{x^2+x+1}{1-x-x^2}\right)dx$ is equal to
Step-by-Step Solution
Key Concept: Integration by parts is applied to integrals of the form $\int xe^x dx$ using the formula $\int u \, dv = uv - \int v \, du$.
To find $I_2 = \int_0^1 2xe^x dx$, we use integration by parts with $u = 2x$ and $dv = e^x dx$. This gives $du = 2dx$ and $v = e^x$. Applying integration by parts: $I_2 = [2xe^x]_0^1 - \int_0^1 2e^x dx = 2e - 0 - 2[e^x]_0^1 = 2e - 2(e-1) = e$.
Correct Answer: e