Question

Consider $\int \frac{3x^4+2x^2+1}{\sqrt{x^4+x^2+1}}dx=f\left(x\right)$ . If $f\left(1\right)=\sqrt{3}$ , then ${\left(f\left(2\right)\right)}^2$ is equal to

Answer 1

Answer: 84

$\int \frac{3x^4+2x^2+1}{\sqrt{x^4+x^2+1}}dx=f\left(x\right)$ Multiplying the numerator and denominator by $x$ in the given integral, we get,
$f\left(x\right)=\int \frac{3x^5+2x^3+x}{\sqrt{x^6+x^4+x^2}}dx$
Now substitute $x^6+x^4+x^2=t^2$
$\Rightarrow 2\left(3x^5+2x^3+x\right)dx=2tdt$
So, $f\left(x\right)=\int \frac{tdt}{t}=\sqrt{x^6+x^4+x^2}+C$
As $f\left(1\right)=\sqrt{3}\Rightarrow C=0$
Hence, $f\left(2\right)=\sqrt{64+16+4}\Rightarrow {\left(f\left(2\right)\right)}^2=84$