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Question: Evaluate \(\int_{}^{}{x^{- 2/3}(1 + x^{2/3})^{- 1}dx}\)...

Evaluate ∫x−2/3(1+x2/3)−1dx\int_{}^{}{x^{- 2/3}(1 + x^{2/3})^{- 1}dx}

A

3tan⁡−1(x1/3)+c3\tan^{- 1}(x^{1/3}) + c

B

3tan⁡−1x+c3\tan^{- 1}x + c

C

3tan⁡−1(x2/3)+c3\tan^{- 1}(x^{2/3}) + c

D

None of these

Answer

3tan⁡−1(x1/3)+c3\tan^{- 1}(x^{1/3}) + c

Explanation

Solution

If we substitute x=t3x = t^{3} (as we knowP∈P \in negative integer)

∴Letx=tk,\therefore Letx = t^{k}, where k is L.C.M. of denominator m and n.

∴x=t3⇒dx=3t2dt\therefore x = t^{3} \Rightarrow dx = 3t^{2}dt

orI=∫3t2dtt2(1+t2)=3∫dtt2+1=3tan⁡−1(t)+c⇒I=3tan⁡−1(x1/3)+cI = \int_{}^{}{\frac{3t^{2}dt}{t^{2}(1 + t^{2})} = 3\int_{}^{}{\frac{dt}{t^{2} + 1} = 3\tan^{- 1}(t) + c}} \Rightarrow I = 3\tan^{- 1}(x^{1/3}) + c.