Question

$\int \frac{dx}{x^2\sqrt{16-x^2}}$ has the value equal to

• A. $-\frac{1}{4}ta{n}^{-1}\left(sec\right(\frac{x}{4}\left)\right)+c$
• B. $\frac{1}{4}ta{n}^{-1}\left(sec\right(\frac{x}{4}\left)\right)+c$
• C. $-\frac{\sqrt{16-x^2}}{16x}+c$
• D. $\frac{\sqrt{16+x^2}}{16x}+c$

Answer 1

The correct option is C

$-\frac{\sqrt{16-x^2}}{16x}+c$

$I=\int \frac{1}{x^2\sqrt{16-x^2}}dx$

Put $x=\frac{1}{t};$

$dx=-\frac{1}{t^2}dt\therefore I=\int \frac{-\frac{1}{t^2}dt}{\frac{1}{t^2}\sqrt{16-\frac{1}{t^2}}}=\int \frac{-tdt}{\sqrt{16t^2-1}}$

Let $16t^2-1=u^2;32tdt=2udu;$

$tdt=\frac{u}{16}du\therefore I=-\frac{1}{16}\int \frac{udu}{u}=-\frac{u}{16}+c=-\frac{\sqrt{16-x^2}}{16x}+C$