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QJEEM24Ma8ac6: JEE Main 2024 Mathematics - Class 12 JEEM24Ma8ac6 Solution

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

Multiple Choice Options

  • A log⁡e(∣tan⁡−1(x3+1x3)∣)1/3+C\log _{\mathrm{e}}\left(\left|\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right|\right)^{1 / 3}+\mathrm{C}
  • B log⁡e(∣tan⁡−1(x3+1x3)∣)+C\log _{\mathrm{e}}\left(\left|\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right|\right)+\mathrm{C}
  • C log⁡e(∣tan⁡−1(x3+1x3)∣)1/2+C\log _{\mathrm{e}}\left(\left|\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right|\right)^{1 / 2}+\mathrm{C}
  • D log⁡e(∣tan⁡−1(x3+1x3)∣)3+C\log _{\mathrm{e}}\left(\left|\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right|\right)^3+\mathrm{C}

Step-by-Step Verified Solution

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

amp; Let tan⁡−1(x3+1x3)=tamp;⇒11+(x3+1x3)2⋅(3x2−3x4)dx=dtamp;⇒x6x12+3x6+1⋅3x6−3x4dx=dtamp;I=13∫dtt=13ln⁡∣t∣+Camp;I=13ln⁡∣tan⁡−1(x3+1x3)∣+Camp;I=ln⁡∣tan⁡−1(x3+1x3)∣1/3+C\begin{aligned}& \text { Let } \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)=t \\& \Rightarrow \frac{1}{1+\left(x^3+\frac{1}{x^3}\right)^2} \cdot\left(3 x^2-\frac{3}{x^4}\right) d x=d t \\& \Rightarrow \frac{x^6}{x^{12}+3 x^6+1} \cdot \frac{3 x^6-3}{x^4} d x=d t \\& I=\frac{1}{3} \int \frac{d t}{t}=\frac{1}{3} \ln |t|+C \\& I=\frac{1}{3} \ln \left|\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right|+C \\& I=\ln \left|\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right|^{1 / 3}+C\end{aligned}

Hence option (1) is correct.

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