Integral Calculus Question 323
Question: $ \int_{{}}^{{}}{\frac{x^{5}\ dx}{\sqrt{(1+x^{3})}}=} $
[IIT 1975]
Options:
A) $ \frac{2}{3}\sqrt{(1+x^{3})}(x^{3}+2) $
B) $ \frac{2}{9}\sqrt{(1+x^{3})}(x^{3}-4) $
C) $ \frac{2}{9}\sqrt{(1+x^{3})}(x^{3}+4) $
D) $ \frac{2}{9}\sqrt{(1+x^{3})}(x^{3}-2) $
Show Answer
Answer:
Correct Answer: D
Solution:
Here $ x^{5}=x^{3}x^{2} $ and differential coefficient of $ x^{3} $ is $ 3{x^{2.}} $ In order to remove fractional powers, we put $ 1+x^{3}=t^{2}\Rightarrow 3x^{2}dx=2t.,dt,; $ Also $ x^{3}=t^{2}-1 $ $ I=\int_{{}}^{{}}{\frac{(t^{2}-1)}{t}( \frac{2}{3}t,dt )=\frac{2}{3}\int_{{}}^{{}}{(t^{2}-1),dt}} $ $ =\frac{2}{3}( \frac{t^{3}}{3}-t )=\frac{2}{9}t,(t^{2}-3) $ = $ \frac{2}{9}\sqrt{(1+x^{3})},(x^{3}-2) $
BETA
