We have to find the following integral

\(\displaystyle∫{\left(\frac{{{2}{x}^{{3}}}}{{{1}+{x}^{{4}}}}\right)}{\left.{d}{x}\right.}\)

Notice that

\(\displaystyle∫{\left(\frac{{{2}{x}^{{3}}}}{{{1}+{x}^{{4}}}}\right)}{\left.{d}{x}\right.}={2}∫{\left(\frac{{{x}^{{3}}}}{{{1}+{x}^{{4}}}}\right)}{\left.{d}{x}\right.}\)

Take \(\displaystyle{u}={1}+{x}^{{4}}\). It follows PSKdu=4(x^3)dx->(x^3)dx=(1/4)du =2∫(1/4u)du =2*1/4ln|u| =1/2ln|1+x^4|ZSK

Therefore we get

\(\displaystyle∫{\left(\frac{{{2}{x}^{{3}}}}{{{1}+{x}^{{4}}}}\right)}{\left.{d}{x}\right.}=\frac{{1}}{{2}}{\ln}{\left|{1}+{x}^{{4}}\right|}+{C}\)

\(\displaystyle∫{\left(\frac{{{2}{x}^{{3}}}}{{{1}+{x}^{{4}}}}\right)}{\left.{d}{x}\right.}\)

Notice that

\(\displaystyle∫{\left(\frac{{{2}{x}^{{3}}}}{{{1}+{x}^{{4}}}}\right)}{\left.{d}{x}\right.}={2}∫{\left(\frac{{{x}^{{3}}}}{{{1}+{x}^{{4}}}}\right)}{\left.{d}{x}\right.}\)

Take \(\displaystyle{u}={1}+{x}^{{4}}\). It follows PSKdu=4(x^3)dx->(x^3)dx=(1/4)du =2∫(1/4u)du =2*1/4ln|u| =1/2ln|1+x^4|ZSK

Therefore we get

\(\displaystyle∫{\left(\frac{{{2}{x}^{{3}}}}{{{1}+{x}^{{4}}}}\right)}{\left.{d}{x}\right.}=\frac{{1}}{{2}}{\ln}{\left|{1}+{x}^{{4}}\right|}+{C}\)