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NVAMediumJEE 2026Indefinite Integrals

JEE Mathematics 2026 Question with Solution

If ∫014cot⁡−1(1−2x+4x2) dx=atan⁡−1(2)−bln⁡(5)\int_0^1 4 \cot^{-1}(1 - 2x + 4x^2) \, dx = a \tan^{-1}(2) - b \ln(5), where a,b∈Na, b \in \mathbb{N}, then (2a+b)(2a + b) is equal to _____.

Answer

Correct answer:9

Step-by-step solution

Standard Method

Given:

I=∫014cot⁡−1(1−2x+4x2) dxI=\int_0^1 4\cot^{-1}(1-2x+4x^2)\,dx

and

I=atan⁡−1(2)−bln⁡(5)I=a\tan^{-1}(2)-b\ln(5)

Find: the value of (2a+b)(2a+b).

Use

cot⁡−1z=tan⁡−1(1z)\cot^{-1}z=\tan^{-1}\left(\frac{1}{z}\right)

and the identity

tan⁡−1A−tan⁡−1B=tan⁡−1(A−B1+AB).\tan^{-1}A-\tan^{-1}B=\tan^{-1}\left(\frac{A-B}{1+AB}\right).

Now,

cot⁡−1(1−2x+4x2)=tan⁡−1(11+2x(2x−1)).\cot^{-1}(1-2x+4x^2)=\tan^{-1}\left(\frac{1}{1+2x(2x-1)}\right).

Since

2x−(2x−1)=1,2x-(2x-1)=1,

we can write

tan⁡−1(2x−(2x−1)1+(2x)(2x−1))=tan⁡−1(2x)−tan⁡−1(2x−1).\tan^{-1}\left(\frac{2x-(2x-1)}{1+(2x)(2x-1)}\right)=\tan^{-1}(2x)-\tan^{-1}(2x-1).

Therefore,

I=4∫01[tan⁡−1(2x)−tan⁡−1(2x−1)] dx.I=4\int_0^1 [\tan^{-1}(2x)-\tan^{-1}(2x-1)]\,dx.

For the second integral, let

t=2x−1,t=2x-1,

so that

dx=dt2.dx=\frac{dt}{2}.

Then

∫01tan⁡−1(2x−1) dx=12∫−11tan⁡−1(t) dt=0,\int_0^1 \tan^{-1}(2x-1)\,dx=\frac{1}{2}\int_{-1}^{1}\tan^{-1}(t)\,dt=0,

because tan⁡−1(t)\tan^{-1}(t) is an odd function over the symmetric interval [−1,1][-1,1].

Hence,

I=4∫01tan⁡−1(2x) dx.I=4\int_0^1 \tan^{-1}(2x)\,dx.

Let

u=2x,u=2x,

so that

dx=du2.dx=\frac{du}{2}.

Thus,

I=2∫02tan⁡−1u du.I=2\int_0^2 \tan^{-1}u\,du.

Using

∫tan⁡−1u du=utan⁡−1u−12ln⁡(1+u2),\int \tan^{-1}u\,du=u\tan^{-1}u-\frac{1}{2}\ln(1+u^2),

we get

I=2[utan⁡−1u−12ln⁡(1+u2)]02.I=2\left[u\tan^{-1}u-\frac{1}{2}\ln(1+u^2)\right]_0^2.

Now evaluate:

I=2[(2tan⁡−12−12ln⁡5)−0]=4tan⁡−12−ln⁡5.I=2\left[(2\tan^{-1}2-\frac{1}{2}\ln5)-0\right]=4\tan^{-1}2-\ln5.

Comparing with

I=atan⁡−1(2)−bln⁡(5),I=a\tan^{-1}(2)-b\ln(5),

we obtain

a=4,b=1.a=4,\quad b=1.

Therefore,

2a+b=2(4)+1=9.2a+b=2(4)+1=9.

So the required numerical value is 99.

Using arctangent difference pattern

Given: the integrand contains 1−2x+4x2=1+(2x)(2x−1)1-2x+4x^2=1+(2x)(2x-1).

Find: convert the inverse cotangent into a simpler difference of inverse tangents.

Notice that the denominator matches the 1+AB1+AB structure with

A=2x,B=2x−1.A=2x, \qquad B=2x-1.

Also,

A−B=2x−(2x−1)=1.A-B=2x-(2x-1)=1.

Hence

A−B1+AB=11+(2x)(2x−1)=11−2x+4x2.\frac{A-B}{1+AB}=\frac{1}{1+(2x)(2x-1)}=\frac{1}{1-2x+4x^2}.

So,

cot⁡−1(1−2x+4x2)=tan⁡−1(11−2x+4x2)=tan⁡−1(2x)−tan⁡−1(2x−1).\cot^{-1}(1-2x+4x^2)=\tan^{-1}\left(\frac{1}{1-2x+4x^2}\right)=\tan^{-1}(2x)-\tan^{-1}(2x-1).

This reduces the original integral to standard integrals, after which the odd-function part vanishes and the remaining part gives

I=4tan⁡−12−ln⁡5.I=4\tan^{-1}2-\ln5.

Thus a=4a=4 and b=1b=1, so

2a+b=9.2a+b=9.

Common mistakes

  • Using the identity for tan⁡−1(A−B)\tan^{-1}(A-B) directly is wrong because the required formula is for tan⁡−1A−tan⁡−1B\tan^{-1}A-\tan^{-1}B with the denominator 1+AB1+AB. First match the expression to A−B1+AB\frac{A-B}{1+AB}, then split it.

  • Missing that ∫−11tan⁡−1(t) dt=0\int_{-1}^{1}\tan^{-1}(t)\,dt=0 is a conceptual error. Since tan⁡−1(t)\tan^{-1}(t) is odd and the interval is symmetric, that entire part cancels immediately.

  • While substituting u=2xu=2x or t=2x−1t=2x-1, forgetting the factor in dxdx gives a wrong coefficient. Always replace dxdx correctly: dx=du2dx=\frac{du}{2} or dx=dt2dx=\frac{dt}{2}.

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