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MCQMediumJEE 2024Indefinite Integrals

JEE Mathematics 2024 Question with Solution

The integral ∫x8−x2(x12+3x6+1)arctan⁡(x3+1x3) dx\int \frac{x^8 - x^2}{(x^{12} + 3x^6 + 1) \arctan\left(x^3 + \frac{1}{x^3}\right)} \, dx is equal to:

  • A

    log⁡(arctan⁡(x3+1x3)1/3)+C\log\left(\arctan\left(x^3 + \frac{1}{x^3}\right)^{1/3}\right) + C

  • B

    log⁡(arctan⁡(x3+1x3)1/2)+C\log\left(\arctan\left(x^3 + \frac{1}{x^3}\right)^{1/2}\right) + C

  • C

    log⁡(arctan⁡(x3+1x3))+C\log\left(\arctan\left(x^3 + \frac{1}{x^3}\right)\right) + C

  • D

    log⁡(arctan⁡(x3+1x3)3)+C\log\left(\arctan\left(x^3 + \frac{1}{x^3}\right)^3\right) + C

Answer

Correct answer:A

Step-by-step solution

Substitution Method

Given:

I=∫x8−x2(x12+3x6+1) arctan⁡(x3+1x3) dxI = \int \frac{x^8 - x^2}{(x^{12} + 3x^6 + 1) \, \arctan\left(x^3 + \frac{1}{x^3}\right)} \, dx

Find: The value of the integral.

Take the substitution

t=arctan⁡(x3+1x3)t = \arctan\left(x^3 + \frac{1}{x^3}\right)

Then

dt=11+(x3+1x3)2(3x2−3x4)dxdt = \frac{1}{1 + \left(x^3 + \frac{1}{x^3}\right)^2} \left(3x^2 - \frac{3}{x^4}\right) dx

Simplifying,

dt=11+(x3+1x3)2⋅3x6−3x4 dxdt = \frac{1}{1 + \left(x^3 + \frac{1}{x^3}\right)^2} \cdot \frac{3x^6 - 3}{x^4} \, dx

Using

1+(x3+1x3)2=x12+3x6+1x61 + \left(x^3 + \frac{1}{x^3}\right)^2 = \frac{x^{12} + 3x^6 + 1}{x^6}

we get

dt=3(x8−x2)x12+3x6+1 dxdt = \frac{3(x^8 - x^2)}{x^{12} + 3x^6 + 1} \, dx

Therefore,

x8−x2x12+3x6+1 dx=13 dt\frac{x^8 - x^2}{x^{12} + 3x^6 + 1} \, dx = \frac{1}{3} \, dt

So the integral becomes

I=13∫dttI = \frac{1}{3} \int \frac{dt}{t}

Now integrate:

I=13ln⁡∣t∣+CI = \frac{1}{3} \ln|t| + C

Substituting back,

I=13ln⁡∣arctan⁡(x3+1x3)∣+CI = \frac{1}{3} \ln\left|\arctan\left(x^3 + \frac{1}{x^3}\right)\right| + C

This can be written as

I=ln⁡(arctan⁡(x3+1x3))1/3+CI = \ln\left(\arctan\left(x^3 + \frac{1}{x^3}\right)\right)^{1/3} + C

Therefore, the correct option is A.

Why the numerator matches the derivative

The key observation is that

ddx(x3+1x3)=3x2−3x4=3(x6−1)x4\frac{d}{dx}\left(x^3 + \frac{1}{x^3}\right) = 3x^2 - \frac{3}{x^4} = \frac{3(x^6 - 1)}{x^4}

Also,

x8−x2=x2(x6−1)x^8 - x^2 = x^2(x^6 - 1)

Further,

1+(x3+1x3)2=1+x6+2+1x6=x6+3+1x6=x12+3x6+1x61 + \left(x^3 + \frac{1}{x^3}\right)^2 = 1 + x^6 + 2 + \frac{1}{x^6} = x^6 + 3 + \frac{1}{x^6} = \frac{x^{12} + 3x^6 + 1}{x^6}

Hence,

11+(x3+1x3)2(3x2−3x4)=x6x12+3x6+1⋅3(x6−1)x4=3x2(x6−1)x12+3x6+1=3(x8−x2)x12+3x6+1\frac{1}{1 + \left(x^3 + \frac{1}{x^3}\right)^2} \left(3x^2 - \frac{3}{x^4}\right) = \frac{x^6}{x^{12} + 3x^6 + 1} \cdot \frac{3(x^6 - 1)}{x^4} = \frac{3x^2(x^6 - 1)}{x^{12} + 3x^6 + 1} = \frac{3(x^8 - x^2)}{x^{12} + 3x^6 + 1}

That is exactly the integrand numerator up to the factor 33, which gives the factor 13\frac{1}{3} in the final answer.

Common mistakes

  • A common mistake is differentiating arctan⁡(x3+1x3)\arctan\left(x^3 + \frac{1}{x^3}\right) without using the chain rule. This misses the factor 11+u2\frac{1}{1+u^2}. Always use ddx(arctan⁡u)=u′1+u2\frac{d}{dx}(\arctan u)=\frac{u'}{1+u^2}.

  • Students often simplify 1+(x3+1x3)21+\left(x^3+\frac{1}{x^3}\right)^2 incorrectly. Expanding it properly gives x6+3+1x6=x12+3x6+1x6x^6+3+\frac{1}{x^6} = \frac{x^{12}+3x^6+1}{x^6}, not just x6+1x6x^6+\frac{1}{x^6}.

  • Another mistake is forgetting the factor 13\frac{1}{3} after substitution. Since dt=3(x8−x2)x12+3x6+1 dxdt = \frac{3(x^8-x^2)}{x^{12}+3x^6+1} \, dx, the integrand becomes 13dtt\frac{1}{3}\frac{dt}{t}, not dtt\frac{dt}{t}.

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