Integral Calculus Question 294
Question: $ \int_{{}}^{{}}{x\sqrt{1+x^{2}}}\ dx= $
[MP PET 1989]
Options:
A) $ \frac{1+2x^{2}}{\sqrt{1+x^{2}}}+c $
B) $ \sqrt{1+x^{2}}+c $
C) $ 3{{(1+x^{2})}^{3/2}}+c $
D) $ \frac{1}{3}{{(1+x^{2})}^{3/2}}+c $
Show Answer
Answer:
Correct Answer: D
Solution:
Put $ 1+x^{2}=t\Rightarrow x,dx=\frac{dt}{2} $ It reduces to $ \frac{1}{2}\int_{{}}^{{}}{{t^{1/2}}dt}=\frac{1}{2}\times \frac{{t^{3/2}}}{3/2}=\frac{1}{3}{{(1+x^{2})}^{3/2}}+c. $
BETA
