Question

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

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

Answer 1

The correct option is B $\frac{\sqrt{x^4+x^2+1}}{x}+c$

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

$=\int \frac{x^4-1}{x^3\sqrt{x^2+1+\frac{1}{x^2}}}dx$ (dividing and multiplying the denominator by $x$)

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

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

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

The integral becomes

$\int \frac{dt}{2\sqrt{1+t}}=(t+1{)}^{1/2}+c$

Substituting back the value of $t$, we get integral

$=\frac{\sqrt{x^4+x^2+1}}{x}+c$