$int (1+x^2)/(1+7x^2+x^4)dx$ ?

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Answer 1 · 2017-06-06T13:54:59

$1/3arc tan{(x^2-1)/(3x)}+C.$

Suppose that, $I=int(1+x^2)/(1+7x^2+x^4)dx.$

$:. I=int{x^2(1/x^2+1)}/{x^2(1/x^2+7+x^2)}dx.$

$=int(1+1/x^2)/(x^2+1/x^2+7)dx$

We subst. $u=x-1/x rArr du=(1-(-1/x^2))dx=(1+1/x^2)dx.$

Also, $x^2+1/x^2+7=(x-1/x)^2+9=u^2+9.$

Therefore, $I=int1/(u^2+3^2)du,$

$=1/3arc tan(u/3)$

$rArr I=1/3arc tan{(x-1/x)/3}=1/3arc tan{(x^2-1)/(3x)}+C.$

Enjoy Maths.!

int (1+x^2)/(1+7x^2+x^4)dxint (1+x^2)/(1+7x^2+x^4)dx?