Question

Evaluate: $\int \frac{x^4-2}{x^2\sqrt{x^4+x^2+2}}dx$

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

Answer 1

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

Let $I=\int \frac{x^4-2}{x^2\sqrt{x^4+x^2+2}}dx=\int \frac{(x^2-\sqrt{2})(x^2+\sqrt{2})}{x^3\sqrt{x^2+\frac{2}{x^2}+1}}$
$=\int \frac{(x+\frac{\sqrt{2}}{x})(x-\frac{\sqrt{2}}{x})}{\sqrt{{(x+\frac{\sqrt{2}}{x})}^{2}2\sqrt{2}-1}}$
Substitute ${(x+\frac{2\sqrt{2}}{x})}^2-2\sqrt{2}-1=t\Rightarrow 2(x+\frac{\sqrt{2}}{x})(1-\frac{\sqrt{2}}{x^2})dx=dt$
$\therefore I\int \frac{dt}{2\sqrt{t}}=\sqrt{t}+c=\sqrt{x^2+\frac{1}{x^2}+2+c}=\sqrt{\frac{x^4+x^2+1}{x}+c}$