Question

$\int \frac{\sqrt{4+x^2}}{x^6}dx=\frac{A{(4+x^2)}^{\frac{3}{2}}(B{x}^2-6)}{x^5}+C$

• A. $A=\frac{1}{120}$
• B. $B=1$
• C. $A=-\frac{1}{120}$
• D. $B=-1$

Answer 1

Answer: (A), (B)

$B=1$
$A=\frac{1}{120}$

Let $I=\int \frac{\sqrt{x^2+4}}{x^6}dx$

Substituting $x=2\tan t\Rightarrow dx=2{\sec }^{2}tdt$
$I=2\int \frac{1}{32}{\cot }^{3}t{\csc }^{3}tdt=\frac{1}{16}\int \cot t{\csc }^{3}t({\csc }^{2}t-1)dt$

Again substituting $u=\csc t\Rightarrow du=-\cot t\csc tdt$

$I=-\frac{1}{16}\int u^2(u^2-1)du=-\frac{1}{16}\int (u^4-u^2)du$

$=\frac{u^3}{48}-\frac{u^5}{80}+c=\frac{{\csc }^{3}t}{48}-\frac{{\csc }^{5}t}{80}+c$

$=\frac{{(x^2+4)}^{3/2}}{48x^3}-\frac{{(x^2+4)}^{5/2}}{80x^5}+c=\frac{\sqrt{x^2+4}(x^4-2x^2-24)}{120x^5}+c$

$=\frac{{(x^2+4)}^{3/2}(x^2-6)}{120x^5}+c$

Hence, $A=\frac{1}{120}$ and $B=1$.