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: Convert
\(\cot^{-1}(1+x^2-x) = \tan^{-1}\frac{1}{1-x+x^2}\), because the argument is positive on \([0,1]\).

Step 2: Split
\(\frac{1}{1-x(1-x)} = \frac{x+(1-x)}{1-x(1-x)}\), which is the tangent of a sum. So the term equals \(\tan^{-1}x + \tan^{-1}(1-x)\).

Step 3: Use the property
Since \(\int_0^1\tan^{-1}(1-x)dx = \int_0^1\tan^{-1}x\,dx\), the integral is \(2\int_0^1\tan^{-1}x\,dx\).

Step 4: Integrate by parts
\(\int_0^1\tan^{-1}x\,dx = \left[x\tan^{-1}x\right]_0^1 - \int_0^1\frac{x}{1+x^2}dx = \frac\pi4 - \frac12\ln2\).

Step 5: Result
Double: \(\frac\pi2 - \ln2\). Option (A).

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