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Question: If I = \(\int_{}^{}\frac{dx}{x\sqrt{1 - x^{3}}}\), then I equals:...

If I = ∫dxx1−x3\int_{}^{}\frac{dx}{x\sqrt{1 - x^{3}}}, then I equals:

A

13\frac{1}{3}log∣1−x3−11−x3+1∣\left| \frac{\sqrt{1 - x^{3}} - 1}{\sqrt{1 - x^{3}} + 1} \right| + C

B

13\frac{1}{3}log ∣1−x3+11−x3−1∣\left| \frac{\sqrt{1 - x^{3}} + 1}{\sqrt{1 - x^{3}} - 1} \right| + C

C

23\frac{2}{3}log |1 – x3| + C

D

13\frac{1}{3}log ∣x3/2+1−x3∣\left| x^{3/2} + \sqrt{1 - x^{3}} \right| + C

Answer

13\frac{1}{3}log ∣1−x3+11−x3−1∣\left| \frac{\sqrt{1 - x^{3}} + 1}{\sqrt{1 - x^{3}} - 1} \right| + C

Explanation

Solution

Write I = ∫x2dxx31−x3\int_{}^{}\frac{x^{2}dx}{x^{3}\sqrt{1 - x^{3}}}and 1 –x3 = t2,

so that –3x2 dx = 2t dt and

I = = – 23\frac { 2 } { 3 } ∫dt1−t2\int_{}^{}\frac{dt}{1 - t^{2}}

= – 23\frac { 2 } { 3 } (12)\left( \frac{1}{2} \right)log ∣1−t1+t∣\left| \frac{1 - t}{1 + t} \right|+ C = 3 log ∣1−x3+11−x3−1∣\left| \frac{\sqrt{1 - x^{3}} + 1}{\sqrt{1 - x^{3}} - 1} \right|+ C