Question

$\int \frac{(1+x^2)dx}{(1-x^2)\sqrt{1+x^2+x^4}}=$

• A. $I=-\frac{1}{2\sqrt{3}}\log \left|\frac{\sqrt{x^2+\frac{1}{x^2}+1}-\sqrt{5}}{\sqrt{x^2+\frac{1}{x^2}+1}+\sqrt{5}}\right|+C$
• B. $I=-\frac{1}{4\sqrt{3}}\log \left|\frac{\sqrt{x^2+\frac{1}{x^2}+1}-\sqrt{3}}{\sqrt{2x^2+\frac{1}{x^2}+1}+\sqrt{3}}\right|+C$
• C. $I=-\frac{1}{2\sqrt{3}}\log \left|\frac{\sqrt{2x^2+\frac{1}{2x^2}+1}-\sqrt{3}}{\sqrt{2x^2+\frac{1}{2x^2}+1}+\sqrt{3}}\right|+C$
• D. $I=-\frac{1}{2\sqrt{3}}\log \left|\frac{\sqrt{x^2+\frac{1}{x^2}+1}-\sqrt{3}}{\sqrt{x^2+\frac{1}{x^2}+1}+\sqrt{3}}\right|+C$

Answer 1

Answer: (D)

$I=-\frac{1}{2\sqrt{3}}\log \left|\frac{\sqrt{x^2+\frac{1}{x^2}+1}-\sqrt{3}}{\sqrt{x^2+\frac{1}{x^2}+1}+\sqrt{3}}\right|+C$

Let $I=\int \frac{(1+x^2)dx}{(1-x^2)\sqrt{1+x^2+x^4}}=\int \frac{x^2(1+\frac{1}{x^2})dx}{x(\frac{1}{x}-x)x\sqrt{\frac{1}{x^2}+1+x^2}}$.
$=-\int \frac{(1+\frac{1}{x^2})dx}{(1-\frac{1}{x})\sqrt{{(x-\frac{1}{x})}^2+3}}$
Put $x-\frac{1}{x}=t\Rightarrow -\int \frac{dt}{t\sqrt{t^2+3}}$ as $(1+\frac{1}{x^2})dx=dt$
Again, put $t^2+3=s^2⟹2tdt=2sds$
$=-\int \frac{sds}{s(s^2-3)}=-\int \frac{ds}{s^2-{\left(\sqrt{3}\right)}^2}$

$=-\frac{1}{2\sqrt{3}}\log \left|\frac{s-\sqrt{3}}{s+\sqrt{3}}\right|+C$
$=-\frac{1}{2\sqrt{3}}\log \left|\frac{\sqrt{t^2+3}-\sqrt{3}}{\sqrt{t^2+3}+\sqrt{3}}\right|+C$

$=-\frac{1}{2\sqrt{3}}\log \left|\frac{\sqrt{{(x-\frac{1}{x})}^2+3}-\sqrt{3}}{\sqrt{{(x-\frac{1}{x})}^2+3}+\sqrt{3}}\right|+C$
$I=-\frac{1}{2\sqrt{3}}\log \left|\frac{\sqrt{x^2+\frac{1}{x^2}+1}-\sqrt{3}}{\sqrt{x^2+\frac{1}{x^2}+1}+\sqrt{3}}\right|+C$

Hence, option D.