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