∫((2x^3)/(1+x^4))dx

Question
Exponents and radicals
asked 2020-10-31
\(\displaystyle∫{\left(\frac{{{2}{x}^{{3}}}}{{{1}+{x}^{{4}}}}\right)}{\left.{d}{x}\right.}\)

Answers (1)

2020-11-01
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
\(\displaystyle∫{\left(\frac{{{2}{x}^{{3}}}}{{{1}+{x}^{{4}}}}\right)}{\left.{d}{x}\right.}=\frac{{1}}{{2}}{\ln}{\left|{1}+{x}^{{4}}\right|}+{C}\)
0

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