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 $u=(x^4+1)^(1/4)/x$

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

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