A) \[\frac{x}{{{\left( {{a}^{2}}+{{x}^{2}} \right)}^{1/2}}}\]
B) \[\frac{x}{{{a}^{2}}{{\left( {{a}^{2}}+{{x}^{2}} \right)}^{1/2}}}\]
C) \[\frac{1}{{{a}^{2}}{{\left( {{a}^{2}}+{{x}^{2}} \right)}^{1/2}}}\]
D) None of these
Correct Answer: B
Solution :
\[I=\frac{1}{2}\log \left| \frac{1+\tan x}{1-\tan x} \right|+C\] Put \[x=a\tan \theta \,\,\Rightarrow dx=a{{\sec }^{2}}\theta \,d\theta \] \[\therefore \,I=\int{\frac{a{{\sec }^{2}}\theta }{{{({{a}^{2}}+{{a}^{2}}{{\tan }^{2}}\theta )}^{3/2}}}d\theta }=\int{\frac{a{{\sec }^{2}}\theta }{{{a}^{3}}{{({{\sec }^{2}}\theta )}^{3/2}}}d\theta }\] Þ \[I=\frac{1}{{{a}^{2}}}\int{\frac{d\theta }{\sec \theta }}\] \[=\frac{1}{{{a}^{2}}}\int{\cos \theta \,d\theta =\frac{1}{{{a}^{2}}}\sin \theta +c}\] Þ \[I=\frac{x}{{{a}^{2}}{{({{x}^{2}}+{{a}^{2}})}^{1/2}}}+c\].
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