Integral Calculus Question 465
Question: The value of $ \int_{{}}^{{}}{\frac{x^{3}}{\sqrt{1+x^{4}}}\ dx} $ is
[SCRA 1996]
Options:
A) $ {{(1+x^{4})}^{\frac{1}{2}}}+c $
B) $ -{{(1+x^{4})}^{\frac{1}{2}}}+c $
C) $ \frac{1}{2}{{(1+x^{4})}^{\frac{1}{2}}}+c $
D) $ -\frac{1}{2}{{(1+x^{4})}^{\frac{1}{2}}}+c $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{\frac{x^{3}}{\sqrt{1+x^{4}}},dx}=\frac{1}{4}\int_{{}}^{{}}{\frac{4x^{3}}{\sqrt{1+x^{4}}},dx} $ $ (Put,1+x^{4}=t) $ $ =\frac{1}{4}\int_{{}}^{{}}{\frac{dt}{{t^{1/2}}}}=\frac{1}{4}\frac{{t^{-\frac{1}{2}+1}}}{-\frac{1}{2}+1}=\frac{1}{2}\sqrt{t}=\frac{1}{2}\sqrt{1+x^{4}}+c. $
BETA
