A) \[\frac{{{(1+{{x}^{2}})}^{3/2}}}{3}\sqrt{1+{{x}^{2}}}+c\]
B) \[{{x}^{2}}\sqrt{1+{{x}^{2}}}-\frac{1}{3}{{(1+{{x}^{2}})}^{3}}=c\]
C) \[\frac{1}{3}{{x}^{2}}\sqrt{1+{{x}^{2}}}-\frac{2}{3}{{(1+{{x}^{2}})}^{3}}+c\]
D) none of these
Correct Answer: C
Solution :
[C]| \[1+{{x}^{2}}={{t}^{2}}\] |
| \[\Rightarrow I=\int{({{t}^{2}}-1)dt=\frac{{{t}^{3}}}{3}}-t\] |
| \[=\frac{{{(1+{{x}^{2}})}^{3/2}}}{3}-\sqrt{1+{{x}^{2}}}\] |
| \[=\frac{(1+{{x}^{2}})\sqrt{1+{{x}^{2}}}}{3}-\sqrt{1+{{x}^{2}}}(1+{{x}^{2}}-{{x}^{2}})\] |
| \[=\frac{1}{3}{{x}^{2}}\sqrt{1+{{x}^{2}}}-\frac{2}{3}{{(1+{{x}^{2}})}^{3/2}}\] |
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