Answer
$$\int_{0}^{1} \frac{x^{3}+2 x}{x^{4}+4 x^{2}+3} d x=\frac{1}{4} \ln \frac{8}{3}$$
Work Step by Step
Given $$\int_{0}^{1} \frac{x^{3}+2 x}{x^{4}+4 x^{2}+3} d x$$
\begin{array}{l}{\text { Let } u=x^{4}+4 x^{2}+3, \text { so that } d u=\left(4 x^{3}+8 x\right) d x}\\{du=4\left(x^{3}+2 x\right) d x, x=0 \Rightarrow u=3, \text { and } x=1 \Rightarrow u=8} \\ {\text { Then } }\\{\begin{align}I&=\int_{0}^{1} \frac{x^{3}+2 x}{x^{4}+4 x^{2}+3} d x\\&=\int_{3}^{8} \frac{1}{u}\left(\frac{1}{4} d u\right)\\
&=\frac{1}{4}[\ln |u|]_{3}^{8}\\&=\frac{1}{4}(\ln 8-\ln 3)\\
&=\frac{1}{4} \ln \frac{8}{3}\end{align}}\end{array}
