Can someone please show me how to solve this Itô Integral?
$$\begin{align}\int_{1}^{t}\frac{dB_s}{B_s^2 + B_s^4} && \end{align} $$
2026-03-29 06:27:17.1774765637
Solving an Itô Integral
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Set $f(x)=-\frac{1}{x}-\tan^{-1}x$. we have $$f'(x)=\frac{1}{x^2}-\frac{1}{1+x^2}=\frac{1}{x^2+x^4}$$ and $$f''(x)=-\frac{2x+4x^3}{(x^2+x^4)^2}$$ By application of Ito's lemma we have $$f(B_t)=f(B_1)+\int_{1}^{t}f'(B_s)dB_s+\frac{1}{2}\int_{1}^{t}f''(B_s)ds$$ therefore $$-\left(\frac{1}{B_t}+\tan^{-1}(B_t)\right)=-\left(\frac{1}{B_1}+\tan^{-1}(B_1)\right)+\int_{0}^{t}\frac{1}{B_s^2+B_s^4}dB_s-\int_{0}^{t}\frac{B_s+2B_s^3}{(B_s^2+B_s^4)^2}ds$$ In other words $$\int_{0}^{t}\frac{1}{B_s^2+B_s^4}dB_s=\int_{0}^{t}\frac{B_s+2B_s^3}{(B_s^2+B_s^4)^2}ds-\left(\frac{1}{B_t}+\tan^{-1}(B_t)\right)+\left(\frac{1}{B_1}+\tan^{-1}(B_1)\right)$$