The integral integral_0^1 cot^-1(1+x+x^2) dx is equal to:

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q20 – JEE-MAIN 2026

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Home Maths Limits and DerivativesQuestion #259

The integral

\int_{0}^{1} \cot^{- 1} (1 + x + x^{2}) d x

is equal to:

Clue 1 of 3: Trigonometric Identity

Convert the $\cot^{- 1}$ function to $\tan^{- 1}$ using the identity $\cot^{- 1} y = \tan^{- 1} (1 / y)$ for $y > 0$ .

🧠 Full Solution Path

● Video Walkthrough

Step 1: Simplify the integrand using trigonometric identities✦ Active

The integrand is $\cot^{- 1} (1 + x + x^{2})$ . Since $1 + x + x^{2} = (x + 1 / 2)^{2} + 3 / 4 > 0$ for all real $x$ , we can use the identity $\cot^{- 1} y = \tan^{- 1} (1 / y)$ . This transforms the integrand to $\tan^{- 1} (\frac{1}{1 + x + x^{2}})$ . We then use the identity $\tan^{- 1} A - \tan^{- 1} B = \tan^{- 1} (\frac{A - B}{1 + A B})$ . By setting $A = x + 1$ and $B = x$ , we can write $\frac{1}{1 + x + x^{2}} = \frac{(x + 1) - x}{1 + x (x + 1)}$ . Thus, the integrand simplifies to $\tan^{- 1} (x + 1) - \tan^{- 1} (x)$ . The integral becomes:

I = \int_{0}^{1} (\tan^{- 1} (x + 1) - \tan^{- 1} (x)) d x

Step 2: Evaluate the indefinite integral of

\tan^{- 1} u

○ Expand

Using integration by parts, $\int \tan^{- 1} u d u = u \tan^{- 1} u - \int \frac{u}{1 + u^{2}} d u$ . To solve the remaining integral, let $w = 1 + u^{2}$ , so $d w = 2 u d u$ . Then $\int \frac{u}{1 + u^{2}} d u = \frac{1}{2} \int \frac{d w}{w} = \frac{1}{2} \log | w | = \frac{1}{2} \log (1 + u^{2})$ . Therefore, the indefinite integral is:

\int \tan^{- 1} u d u = u \tan^{- 1} u - \frac{1}{2} \log (1 + u^{2})

Let $F (u) = u \tan^{- 1} u - \frac{1}{2} \log (1 + u^{2})$ .

💡 Teacher's Secret Hint

Remember the standard integration by parts formula for inverse trigonometric functions.

Step 3: Apply the limits of integration○ Expand

The definite integral can be evaluated as $I = [F (x + 1) - F (x)]_{0}^{1}$ . This simplifies to $I = (F (2) - F (1)) - (F (1) - F (0)) = F (2) - 2 F (1) + F (0)$ . We calculate the values of $F (u)$ at the limits:

F (2) = 2 \tan^{- 1} (2) - \frac{1}{2} \log (1 + 2^{2}) = 2 \tan^{- 1} (2) - \frac{1}{2} \log (5)

F (1) = 1 \tan^{- 1} (1) - \frac{1}{2} \log (1 + 1^{2}) = \frac{π}{4} - \frac{1}{2} \log (2)

F (0) = 0 \tan^{- 1} (0) - \frac{1}{2} \log (1 + 0^{2}) = 0 - \frac{1}{2} \log (1) = 0

Substitute these values back into the expression for $I$ :

I = (2 \tan^{- 1} (2) - \frac{1}{2} \log (5)) - 2 (\frac{π}{4} - \frac{1}{2} \log (2)) + 0

I = 2 \tan^{- 1} (2) - \frac{1}{2} \log (5) - \frac{π}{2} + \log (2)

I = 2 \tan^{- 1} (2) + \log (2) - \frac{1}{2} \log (5) - \frac{π}{2}

I = 2 \tan^{- 1} (2) + \frac{1}{2} (2 \log (2) - \log (5)) - \frac{π}{2}

I = 2 \tan^{- 1} (2) + \frac{1}{2} (\log (4) - \log (5)) - \frac{π}{2}

I = 2 \tan^{- 1} (2) + \frac{1}{2} \log (\frac{4}{5}) - \frac{π}{2}

Since $\log (\frac{4}{5}) = - \log (\frac{5}{4})$ , the final result is:

I = 2 \tan^{- 1} (2) - \frac{1}{2} \log (\frac{5}{4}) - \frac{π}{2}

This matches option 4.

💡 Teacher's Secret Hint

Carefully evaluate the definite integral using the fundamental theorem of calculus and simplify the logarithmic terms.

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Maths Question 20 – JEE-MAIN 2026

Clue 1 of 3: Trigonometric Identity

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