Q. $\int_{}^{}\frac{dx}{\left( a^{2}{+ x}^{2} \right)^{3/2}}$ is equal to

A) $\frac{x}{\left( a^{2} + x^{2} \right)^{1/2}} + c$
B) $\frac{x}{{a^{2}\left( a^{2} + x^{2} \right)}^{1/2}} + c$
C) $\frac{1}{{a^{2}\left( a^{2} + x^{2} \right)}^{1/2}} + c$
D) None of these
Explanation

(b)
Let $I = \int_{}^{}\frac{dx}{\left( a^{2}{+ x}^{2} \right)^{3/2}}$
Put $x = a\tan{\theta \Rightarrow dx = a\sec^{2}{\theta\ d\theta}}$
$\therefore I = \int_{}^{}{\frac{a\sec^{2}\theta}{(a^{2} + a^{2}{\tan^{2}{\theta)}}^{3/2}}d\theta}$
$= \int_{}^{}\frac{a\sec^{2}\theta}{a^{3}({\sec^{2}{\theta)}}^{3/2}}d\theta$
$\Rightarrow I = \frac{1}{a^{2}}\int_{}^{}{\frac{d\theta}{\sec\theta} = \frac{1}{a^{2}}\int_{}^{}{\cos{\theta\ d\theta}}}$
$= \frac{1}{a^{2}}\sec\theta + c$
$\Rightarrow \ \frac{x}{a^{2}\left( x^{2} + a^{2} \right)^{1/2}} + c$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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