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integration of 1/1+x 4

integration of 
1/1+x4

Grade:12

1 Answers

Y RAJYALAKSHMI
45 Points
6 years ago
1/(1+x4) = 1/2[(x2 + 1)/(x4 + 1) – (x2 – 1)/(x4 + 1)]
 
∫(x2 + 1)/(x4 + 1)dx = ∫(1 + 1/x2)/(x2 + 1/x2)dx  (dividing num & den by x2)
put x – 1/x2 = t  => x2 + 1/x2 = t2 + 2  & 1 + 1/x2 dx = dt
∫(1 + 1/x2)/(x2 + 1/x2)dx = ∫dt /(2 + t2)  = 1/srt(2) [tan-1t/sqrt(2)] = 1/srt(2)[tan-1(x2 – 1)/sqrt(2)*x)] …..... (1)
 
∫(x2 – 1)/(x4 + 1)dx = ∫(1 – 1/x2)/(x2 + 1/x2)dx  (dividing num & den by x2)
put x + 1/x2 = t  => x2 + 1/x2 = t2 – 2  & 1 –  1/x2 dx = dt
∫(1 – 1/x2)/(x2 + 1/x2)dx = ∫dt/(t2 – 2) = 1/2*srt(2) [log (t – sqrt(2))/ (t + sqrt(2)) = 1/2*srt(2) [(log (x2 + 1 –  x*sqrt(2))/ (x2 + 1 + x*sqrt(2))] …............ (2)
 
So, ∫1/(1+x4)dx = ½ [(1) + (2)]
 
 
 
 

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