_ 3 ^ 29 [3] (x-2)^ 2 3+ [3] (x-2)^ 2 d x=
Mathematics · JEE Advanced · NTA Exams — Integral Calculus
\[\int_{3}^{29} \frac{\sqrt[3]{(x-2)^{2}}}{3+\sqrt[3]{(x-2)^{2}}} d x=\]
- \(4+\frac{3 \sqrt{3}}{2} \pi\)
- \(2+\frac{3 \sqrt{3}}{2} \pi\)
- \(4+\frac{\sqrt{3}}{2} \pi\)
- \(8+\frac{3 \sqrt{3}}{2} \pi\)
Answer
(D) 8+ 3 3 2
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