∫1x1-x3dx=a log1-x3-11-x3+1+C then the value of a is

I=∫1x1−x3dx=∫x2x31−x3dxLet⁡1−x3=t⇒121−x3x−3x2dx=dt⇒I=23∫1t2−1dt⇒32I=∫1(t+1)(t−1)dt=12∫(t+1)−(t−1)(t+1)(t−1)dt⇒3I=∫1(t−1)dt−∫1(t+1)dt=ln⁡|t−1|−ln⁡|t+1|+cI=13ln⁡1−x3−11−x3+1

∫1x1-x3dx=a log1-x3-11-x3+1+C then the value of a is

I=∫1x1−x3dx=∫x2x31−x3dxLet⁡1−x3=t⇒121−x3x−3x2dx=dt⇒I=23∫1t2−1dt⇒32I=∫1(t+1)(t−1)dt=12∫(t+1)−(t−1)(t+1)(t−1)dt⇒3I=∫1(t−1)dt−∫1(t+1)dt=ln⁡|t−1|−ln⁡|t+1|+cI=13ln⁡1−x3−11−x3+1

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$\int \frac{1}{x \sqrt{1 - x^{3}}} dx = a \log |\frac{\sqrt{1 - x^{3}} - 1}{\sqrt{1 - x^{3}} + 1}| + C$ then the value of a is

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$\frac{2}{3}$

$\frac{1}{3}$

$\frac{- 2}{3}$

$\frac{- 1}{3}$

answer is B.

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Detailed Solution

$\begin{array}{l} I = \int \frac{1}{x \sqrt{1 - x^{3}}} dx = \int \frac{x^{2}}{x^{3} \sqrt{1 - x^{3}}} dx \\ Let \sqrt{1 - x^{3}} = t \\ \Rightarrow \frac{1}{2 \sqrt{1 - x^{3}}} x (- 3 x^{2}) dx = dt \\ \Rightarrow I = \frac{2}{3} \int \frac{1}{t^{2} - 1} dt \\ \Rightarrow \frac{3}{2} I = \int \frac{1}{(t + 1) (t - 1)} dt = \frac{1}{2} \int \frac{(t + 1) - (t - 1)}{(t + 1) (t - 1)} dt \\ \Rightarrow 3 I = \int \frac{1}{(t - 1)} dt - \int \frac{1}{(t + 1)} dt \\ = \ln | t - 1 | - \ln | t + 1 | + c \\ I = \frac{1}{3} \ln |\frac{\sqrt{1 - x^{3}} - 1}{\sqrt{1 - x^{3}} + 1}| \end{array}$