A) \[\log \left| 1+\sqrt{2+{{x}^{2}}} \right|+C\]
B) \[-\log \left| 1+\sqrt{2-{{x}^{2}}} \right|+C\]
C) \[-x\log \left| 1-\sqrt{2-{{x}^{2}}} \right|+C\]
D) \[x\log \left| 1-\sqrt{2+{{x}^{2}}} \right|+C\]
Correct Answer: B
Solution :
\[I=\int_{{}}^{{}}{\frac{x\,dx}{2-{{x}^{2}}+\sqrt{2-{{x}^{2}}}}}\] Put\[t=\sqrt{2-{{x}^{2}}},\frac{dt}{dx}=\frac{1}{2\sqrt{2-{{x}^{2}}}}.(-2x)\] \[\Rightarrow \]\[-tdt=xdx\] \[\therefore \]\[I=\int_{{}}^{{}}{\frac{(-t)dt}{{{t}^{2}}+t}=-\int_{{}}^{{}}{\frac{1}{t+1}dt=-\log |t+1|}}\] \[=-\log \left| \sqrt{2-{{x}^{2}}}+1 \right|+c\]
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