Question

The integral \(\int_{0}^{1} \cot^{-1}(1 + x + x^2) dx\) is equal to:

• A. \(2 \tan^{-1} 2 + \frac{1}{2} \log_e (\frac{5}{4}) + \frac{\pi}{2}\)
• B. \(2 \tan^{-1} 2 + \frac{1}{2} \log_e (\frac{5}{4}) - \frac{\pi}{2}\)
• C. \(2 \tan^{-1} 2 - \frac{1}{2} \log_e (\frac{5}{4}) + \frac{\pi}{2}\)
• D. \(2 \tan^{-1} 2 - \frac{1}{2} \log_e (\frac{5}{4}) - \frac{\pi}{2}\)

Answer 1

The Correct Option is D

Step 1: Convert $\cot^{-1$ into $\tan^{-1}$}
Using the identity \[ \cot^{-1}\theta=\tan^{-1}\left(\frac{1}{\theta}\right) \] we get \[ I=\int_0^1 \tan^{-1}\left(\frac{1}{1+x+x^2}\right)\,dx \] Now write \[ \frac{1}{1+x+x^2} = \frac{(x+1)-x}{1+x(x+1)} \] So, \[ I= \int_0^1 \tan^{-1} \left( \frac{(x+1)-x}{1+x(x+1)} \right)\,dx \] Step 2: Use inverse tangent identity
Using the identity \[ \tan^{-1}\left(\frac{A-B}{1+AB}\right) = \tan^{-1}A-\tan^{-1}B \] Take \[ A=x+1,\qquad B=x \] Then \[ I= \int_0^1 \left[ \tan^{-1}(x+1)-\tan^{-1}x \right]dx \] Split the integral: \[ I= \int_0^1 \tan^{-1}(x+1)\,dx - \int_0^1 \tan^{-1}x\,dx \] Step 3: Change variable in first integral
Let \[ u=x+1 \] Then \[ du=dx \] When $x=0$, $u=1$ When $x=1$, $u=2$ So, \[ I= \int_1^2 \tan^{-1}u\,du - \int_0^1 \tan^{-1}x\,dx \] Step 4: Use standard integral formula
The standard result is \[ \int \tan^{-1}t\,dt = t\tan^{-1}t-\frac{1}{2}\ln(1+t^2) \] Now evaluate both integrals. First integral: \[ \int_1^2 \tan^{-1}u\,du = \left[ u\tan^{-1}u-\frac{1}{2}\ln(1+u^2) \right]_1^2 \] \[ = \left( 2\tan^{-1}2-\frac{1}{2}\ln 5 \right) - \left( \tan^{-1}1-\frac{1}{2}\ln 2 \right) \] Since \[ \tan^{-1}1=\frac{\pi}{4} \] \[ = 2\tan^{-1}2-\frac{1}{2}\ln 5-\frac{\pi}{4}+\frac{1}{2}\ln 2 \] Second integral: \[ \int_0^1 \tan^{-1}x\,dx = \left[ x\tan^{-1}x-\frac{1}{2}\ln(1+x^2) \right]_0^1 \] \[ = \frac{\pi}{4}-\frac{1}{2}\ln 2 \] Step 5: Subtract दोनों integrals
\[ I= \left( 2\tan^{-1}2-\frac{1}{2}\ln 5-\frac{\pi}{4}+\frac{1}{2}\ln 2 \right) - \left( \frac{\pi}{4}-\frac{1}{2}\ln 2 \right) \] \[ = 2\tan^{-1}2-\frac{\pi}{2}-\frac{1}{2}\ln 5+\ln 2 \] Now use \[ \ln 2=\frac{1}{2}\ln 4 \] So, \[ -\frac{1}{2}\ln 5+\ln 2 = -\frac{1}{2}\ln 5+\frac{1}{2}\ln 4 \] \[ = -\frac{1}{2}\ln\left(\frac{5}{4}\right) \] Hence, \[ I= 2\tan^{-1}2 - \frac{1}{2}\log_e\left(\frac{5}{4}\right) - \frac{\pi}{2} \] Final Answer: \[ \boxed{ 2\tan^{-1}2 - \frac{1}{2}\log_e\left(\frac{5}{4}\right) - \frac{\pi}{2} } \] \[ \boxed{\text{Option (D)}} \]