Integral Calculus Question 310
Question: $ \int_{{}}^{{}}{\frac{x^{5}}{\sqrt{1+x^{3}}}dx=} $
[IIT 1985]
Options:
A) $ \frac{2}{9}{{(1+x^{3})}^{3/2}}+c $
B) $ \frac{2}{9}{{(1+x^{3})}^{3/2}}+\frac{2}{3}{{(1+x^{3})}^{1/2}}+c $
C) $ \frac{2}{9}{{(1+x^{3})}^{3/2}}-\frac{2}{3}{{(1+x^{3})}^{1/2}}+c $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Put $ 1+x^{3}=t^{2}\Rightarrow 3x^{2}dx=2t,dt $ and $ x^{3}=t^{2}-1 $
So, $ \int_{{}}^{{}}{\frac{x^{5}}{\sqrt{1+x^{3}}},dx=\int_{{}}^{{}}{\frac{x^{2}.x^{3}}{\sqrt{1+x^{3}}},dx}} $
$ =\frac{2}{3}\int_{{}}^{{}}{\frac{(t^{2}-1),.,t,dt}{t}=\frac{2}{3}\int_{{}}^{{}}{(t^{2}-1),dt=\frac{2}{3}[ \frac{t^{3}}{3}-t ]}}+c $
$ =\frac{2}{3}[ \frac{{{(1+x^{3})}^{3/2}}}{3}-{{(1+x^{3})}^{1/2}} ]+c $ .
BETA
