Question

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

Answer 1

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