A) \[\frac{1+2{{x}^{2}}}{\sqrt{1+{{x}^{2}}}}+c\]
B) \[\sqrt{1+{{x}^{2}}}+c\]
C) \[3{{(1+{{x}^{2}})}^{3/2}}+c\]
D) \[\frac{1}{3}{{(1+{{x}^{2}})}^{3/2}}+c\]
Correct Answer: D
Solution :
Put \[1+{{x}^{2}}=t\Rightarrow x\,dx=\frac{dt}{2}\] It reduces to \[\frac{1}{2}\int_{{}}^{{}}{{{t}^{1/2}}dt}=\frac{1}{2}\times \frac{{{t}^{3/2}}}{3/2}=\frac{1}{3}{{(1+{{x}^{2}})}^{3/2}}+c.\]
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