Question

,

Answer 1

the second is argtan and other let me think

Answer 2

\[\int\limits \frac{ 1 }{ x^4+1 }dx=\frac{ 1 }{ 2 }\int\limits \frac{ 2 }{ x^4+1 }dx=\frac{ 1 }{ 2 }\int\limits \frac{ 1+x^2+1-x^2 }{ x^4+1 } dx\]
\[=\frac{ 1 }{ 2 }\left[ \int\limits \frac{ 1+x^2 }{ x^4+1 }dx-\int\limits \frac{ x^2-1 }{ x^4+1 } dx \right]=\frac{ 1 }{ 2 }\left[ I _{1}+I _{2} \right]+c\]
\[I _{1}=\int\limits \frac{ x^2+1 }{ x^4+1 }dx\]
divide the num. and den. by x^2
\[I _{1}=\int\limits \frac{ 1+\frac{ 1 }{ x^2 } }{ x^2+\frac{ 1 }{ x^2 } }dx\]
put \[x-\frac{ 1 }{ x }=t\]
diff.\[\left( 1+\frac{ 1 }{ x^2 } \right)dx=dt\]
squaring
\[x^2+\frac{ 1 }{ x^2 }-2=t^2,x^2+\frac{ 1 }{ x^2 }=t^2+2\]
?
\[for~I _{2},divide~by~x^2 ,put ~x+\frac{ 1 }{ x }=z,then~proceed~as~before~ and~complete\]