Question

$\text{Let}\int \frac{(x^2-1)dx}{x^3\sqrt{3x^4+2x^2-1}}=f\left(x\right)+c\text{and}\lambda =\underset{x\to \infty }{lim}f\left(x\right)$, then the absolute value of $\frac{\lambda }{\sqrt{3}}$ is (where $c$ is the constant of integration)

Answer 1

$f\left(x\right)=\int \frac{(x^2-1)dx}{x^5\sqrt{3+\frac{2}{x^2}-\frac{1}{x^4}}}f\left(x\right)=\int \frac{(\frac{1}{x^3}-\frac{1}{x^5})dx}{\sqrt{3+\frac{2}{x^2}-\frac{1}{x^4}}}\text{Put},3+\frac{2}{x^2}-\frac{1}{x^4}=t\frac{dt}{dx}=\frac{-4}{x^3}+\frac{4}{x^5}-\frac{1}{4}\int \frac{dt}{\sqrt{t}}=-\frac{1}{2}\sqrt{t}+cf\left(x\right)=-\frac{1}{2}\sqrt{3+\frac{2}{x^2}-\frac{1}{x^4}},\lambda =\underset{x\to \infty }{lim}f\left(x\right)=-\frac{\sqrt{3}}{2}|\frac{\lambda }{\sqrt{3}}|=\frac{1}{2}=0.50$