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what is intigration of 1/(1+x^4)

5 years ago


Answers : (2)


its lower limit is 0 and pper limit is infinity

5 years ago

I = 1/1+x4 dx

 2I = [  2/1+x4]dx

divide nume and denom by  x2

2I = [ 2/x2 /(x2 +1/x2)]dx

2I = (1/x2+1) /[x2+1/x2] dx   +  (1/x2 -1)/[x2+1/x2]dx

2I                            =          (1+1/x2)/[(x-1/x)2+2] dx                 +       (1/x2 -1)/[(x+1/x)2-2] dx

2I                            =                            I1                                     +                     I2

 2I                           =              I1 = 1+1/x2 /[(x-1/x)2+2]dx                  +                  I2 =  (1/x2 -1)/[(x+1/x)2 -2]dx

 2I                           =                put  here   (x-1/x) = t                               &                      put  (x+1/x) = u

 2I                           =                  I1 = dt/[t2+2]                                           +                 I2 = -dt/[u2 -2]

2I                            =        I1 = [tan-1(t/sqrt2)]/sqrt2                      +                 I2 = log(u+sqrt2/u-sqrt2)/2sqrt2      +c

2I                           =       I1 = 1/sqrt2.tan-1(x2-1/xsqrt2)          +         I2 = 1/2sqrt2 .log(x2+1+xsqrt2/x2+1-xsqrt2) + c

this is the required result


5 years ago

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