A) \[-\frac{1}{3}\]
B) \[\frac{1}{3}\]
C) \[-\frac{2}{3}\]
D) \[\frac{2}{3}\]
Correct Answer: B
Solution :
Let\[I=\int{\frac{dx}{x\sqrt{1-{{x}^{3}}}}}=\int{\frac{{{x}^{2}}dx}{{{x}^{3}}\sqrt{1-{{x}^{3}}}}}\] Put \[1-{{x}^{3}}={{t}^{2}}\] \[\Rightarrow \] \[-3{{x}^{2}}dx=2t\text{ }dt\] \[\Rightarrow \] \[{{x}^{2}}dx=-\frac{2t}{3}dt\] \[\therefore \] \[I=-\frac{2}{3}\int{\frac{d\,dt}{(1-{{t}^{2}})\sqrt{{{t}^{2}}}}}\] \[=\frac{2}{3}\int{\frac{dt}{{{t}^{2}}-1}}\] \[=\frac{1}{3}\log \left| \frac{t-1}{t+1} \right|+c\] \[=\frac{1}{3}\log \left| \frac{\sqrt{1-{{x}^{3}}}-1}{\sqrt{1-{{x}^{3}}}+1} \right|+b,\] On comparing with a \[\log \left| \frac{\sqrt{1-{{x}^{3}}}-1}{\sqrt{1-{{x}^{3}}}+1} \right|+b,\] We get \[a=1/3\]
You need to login to perform this action.
You will be redirected in
3 sec
