Integral Calculus – Definite Integration – JEE Main 21 Jan 2026 Shift 2

Solution:

Let the integral be $I$.
$$I = \int_{0}^{1} 4 \cot^{-1}(1-2x+4x^{2}) dx$$

Convert $\cot^{-1}$ to $\tan^{-1}$:
$$I = 4 \int_{0}^{1} \tan^{-1}\left(\frac{1}{1-2x+4x^{2}}\right) dx$$

Factor the denominator to form the identity $\tan^{-1}x – \tan^{-1}y$:
$$I = 4 \int_{0}^{1} \tan^{-1}\left(\frac{1}{1+2x(2x-1)}\right) dx$$

Express the numerator as the difference of the factors in the denominator:
$$I = 4 \int_{0}^{1} \tan^{-1}\left(\frac{2x – (2x-1)}{1+2x(2x-1)}\right) dx$$

$$I = 4 \left[ \int_{0}^{1} \tan^{-1}(2x) dx – \int_{0}^{1} \tan^{-1}(2x-1) dx \right]$$

Let’s evaluate the second integral $I_2 = \int_{0}^{1} \tan^{-1}(2x-1) dx$.
Put $2x-1 = t \Rightarrow 2dx = dt$.
Limits: when $x=0, t=-1$; when $x=1, t=1$.
$$I_2 = \frac{1}{2} \int_{-1}^{1} \tan^{-1}(t) dt$$
Since $\tan^{-1}(t)$ is an odd function, the integral from $-1$ to $1$ is zero.
$$I_2 = 0$$

Now, evaluate the first integral $I = 4 \int_{0}^{1} \tan^{-1}(2x) dx$.
Apply integration by parts with $u = \tan^{-1}(2x)$ and $dv = dx$:
$$I = 4 \left( [x \tan^{-1}(2x)]_{0}^{1} – \int_{0}^{1} x \cdot \frac{2}{1+4x^2} dx \right)$$

$$I = 4 \left( (\tan^{-1}(2) – 0) – \int_{0}^{1} \frac{2x}{1+4x^2} dx \right)$$

For the remaining integral, let $1+4x^2 = u \Rightarrow 8x dx = du \Rightarrow 2x dx = \frac{du}{4}$.
Limits: $x=0 \to u=1$, $x=1 \to u=5$.
$$I = 4 \tan^{-1}(2) – 4 \int_{1}^{5} \frac{1}{4u} du$$

$$I = 4 \tan^{-1}(2) – [\ln |u|]_{1}^{5}$$

$$I = 4 \tan^{-1}(2) – \ln(5)$$

Comparing this with the given form $a \tan^{-1}(2) – b \log_{e}(5)$:
$$a = 4, \quad b = 1$$

We need to find $2a+b$:
$$2a+b = 2(4) + 1 = 9$$

Ans. 9