Question

What is $\int \frac{x^4-1}{x^2\sqrt{x^4+x^2+1}}dx$ equal to?

• A. $\sqrt{\frac{x^4+x^2+1}{x}}+c$
• B. $\sqrt{x^4+2-\frac{1}{x^2}}+c$
• C. $\sqrt{x^2+\frac{1}{x^2}+1}+c$
• D. $\sqrt{\frac{x^4-x^2+1}{x}}+c$

Answer 1

The correct option is C $\sqrt{x^2+\frac{1}{x^2}+1}+c$

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

$=\int \frac{x^4-1}{x^3\sqrt{x^2+1+\frac{1}{x^2}}}dx$

$=\int \frac{x-\frac{1}{x^3}}{\sqrt{x^2+1+\frac{1}{x^2}}}dx$

Put $x^2+1+\frac{1}{x^2}=t$

$\therefore (2x-\frac{2}{x^3})dx=dt$

$\therefore (x-\frac{1}{x^3})dx=\frac{dt}{2}$

$\therefore I=\int \frac{dt}{2\sqrt{t}}$

$=\sqrt{t}+C$

$=\sqrt{x^2+1+\frac{1}{x^2}}+C$

Answer is option (C)