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JEE Main 27-Jan-2024 Paper - Shift 2 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 27-Jan-2024 Paper - Shift 24 Marks
MCQ
The integral $\int \frac{\left(x^8-x^2\right) d x}{\left(x^{12}+3 x^6+1\right) \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)}$ is equal to :
  • A
    $\log _{ e }\left(\left|\tan ^{-1}\left( x ^3+\frac{1}{ x ^3}\right)\right|\right)^{1 / 3}+ C$
  • B
    $\log _{ e }\left(\left|\tan ^{-1}\left( x ^3+\frac{1}{ x ^3}\right)\right|\right)^{1 / 2}+ C$
  • C
    $\log _{ e }\left(\left|\tan ^{-1}\left( x ^3+\frac{1}{ x ^3}\right)\right|\right)+ C$
  • D
    $\log _{ c }\left(\left|\tan ^{-1}\left( x ^3+\frac{1}{ x ^3}\right)\right|\right)^3+ C$
✓

Answer

$I=\int \frac{x^8-x^2}{\left(x^{12}+3 x^6+1\right) \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)} d x$
Let $\tan ^{-1}\left( x ^3+\frac{1}{ x ^3}\right)= t$
$\Rightarrow \frac{1}{1+\left(x^3+\frac{1}{x^3}\right)^2} \cdot\left(3 x^2-\frac{3}{x^4}\right) dx=dt$
$\Rightarrow \frac{x^6}{x^{12}+3 x^6+1} \cdot \frac{3 x^6-3}{x^4} dx=dt$
$I=\frac{1}{3} \int \frac{dt}{t}=\frac{1}{3} \ln |t|+C$
$I=\frac{1}{3} \ln \left|\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right|+C$
$I=\ln \left|\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right|^{1 / 3}+C$
Hence option $(1)$ is correct

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