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MHT CET · Maths · Indefinite Integration

\(\int \frac{4 x^2 \cot ^{-1}\left(x^3\right)}{1+x^6} \mathrm{~d} x=\) (where \(C\) is a constant of integration.)

  1. A \(\frac{-2}{3}\left(\cot ^{-1} X^3\right)+C\)
  2. B \(\frac{-2}{3}\left(\cot ^{-1} X^3\right)^2+C\)
  3. C \(\frac{2}{3}\left(\cot ^{-1} x^3\right)+C\)
  4. D \(\frac{2}{3}\left(\cot ^{-1} x^3\right)^2+C\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{-2}{3}\left(\cot ^{-1} X^3\right)^2+C\)

Step-by-step Solution

Detailed explanation

\(\int \frac{4 x^2 \cot ^{-1}\left(x^3\right)}{1+x^6} \mathrm{~d} x\) let \(\cot ^{-1}\left(x^3\right)=t\)
then \(\frac{-3 x^2}{1+x^6} d x=d t\)
\(\begin{aligned} & =-\frac{4}{3} \int t \mathrm{~d} t \\ & =-\frac{4}{3} \times \frac{t^2}{2}+c=-\frac{2}{3}\left(\cot ^{-1} X^3\right)^2+c\end{aligned}\)