How will you integrate ? #int(dx)/(1+x^4)^(1/4)#

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Answer 1 · 2017-03-29T21:43:00

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Answer 2 · 2017-04-17T10:42:43

#int(dx)/(1+x^4)^(1/4)#

#=int(dx)/(x(1+1/x^4)^(1/4))#

#=int(x^4dx)/(x^5(1+1/x^4)^(1/4))#

Let #(1+1/x^4)=u^4#

Differentiating we get

#-4/x^5dx=4u^3du#

#=>-cancel4/x^5dx=cancel4u^3du#

So Intgral

#I=-int(u^3du)/((u^4-1)u)#

#=-int(u^2du)/((u^2-1)(u^2+1))#

#=-1/2int(((u^2+1)+(u^2-1))du)/((u^2-1)(u^2+1))#

#=-1/2int(du)/(u^2-1)-1/2int(du)/(u^2+1)#

#=-1/4int(((u+1)-(u-1))du)/(u^2-1)-1/2int(du)/(u^2+1)#

#=-1/4int(du)/(u-1)+1/4int(du)/(u+1)-1/2(du)/(u^2+1)#

#=-1/4lnabs(u-1)+1/4lnabs(u+1)-1/2tan^-1u +c#

Inserting #u=(x^4+1)^(1/4)/x#

#I=-1/4lnabs((x^4+1)^(1/4)/x-1)+1/4lnabs((x^4+1)^(1/4)/x+1)-1/2tan^-1((x^4+1)^(1/4)/x)+c#