Question

The value of integral \(\int _0^1cot^{-1}(1+x^2-x)\,dx\) is...

• A. \(\frac{π}{2}-log2\)
• B. \(π-log2\)
• C. \(\frac{π}{4}-log2\)
• D. \(2π-log2\)

Hint:

Write cot inverse as tan inverse of the reciprocal and split it as tan^-1 x + tan^-1(1-x).

Answer 1

The Correct Option is A

Step 1: Identity
For $0\le x\le1$, $\tan^{-1}x+\tan^{-1}(1-x) = \tan^{-1}\frac{1}{1-x+x^2}$ because $x(1-x)<1$.

Step 2: Integrals
By symmetry $x\to1-x$, both parts have the same value. Each equals $\frac\pi4-\frac{\ln2}{2}$.

Step 3: Sum
The total is $\frac\pi2-\ln2$, option (A).

Final Answer:
pi/2 - log 2. \[ \boxed{\text{(A)}\ \frac\pi2-\log2} \]