Integral Calculus Question 117
Question: $ \int_{{}}^{{}}{\frac{x^{3}}{\sqrt{x^{2}+2}}dx=} $
Options:
A) $ \frac{1}{3}{{(x^{2}+2)}^{3/2}}+2{{(x^{2}+2)}^{1/2}}+c $
B) $ \frac{1}{3}{{(x^{2}+2)}^{3/2}}-2{{(x^{2}+2)}^{1/2}}+c $
C) $ \frac{1}{3}{{(x^{2}+2)}^{3/2}}+{{(x^{2}+2)}^{1/2}}+c $
D) $ \frac{1}{3}{{(x^{2}+2)}^{3/2}}-{{(x^{2}+2)}^{1/2}}+c $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{\frac{x^{3}}{\sqrt{x^{2}+2}},dx}=\int_{{}}^{{}}{\frac{x^{2}.,x}{\sqrt{x^{2}+2}},dx} $ Put $ x^{2}+2=t^{2}\Rightarrow x,dx=t,dt $ and $ x^{2}=t^{2}-2, $ then it reduces to $ \int_{{}}^{{}}{\frac{(t^{2}-2)}{t},t,dt}=\int_{{}}^{{}}{(t^{2}-2)dt} $ $ =\frac{t^{3}}{3}-2t+c=\frac{{{(x^{2}+2)}^{3/2}}}{3}-2{{(x^{2}+2)}^{1/2}}+c. $
BETA
