`int 1/(4+4x^2+x^4) dx` Find the indefinite integral

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gsarora17 eNotes educator| Certified Educator

`int1/(4+4x^2+x^4)dx`

Let's rewrite the integrand as :

`=int1/((x^2)^2+2(2)(x^2)+2^2)dx`

`=int1/(x^2+2)^2dx`

Apply integral substitution:`x=sqrt(2)tan(u),u=arctan(x/sqrt(2))`

`dx=sqrt(2)sec^2(u)du`

`=int(sqrt(2)sec^2(u))/((sqrt(2)tan(u))^2+2)^2du`

`=int(sqrt(2)sec^2(u))/(2tan^2(u)+2)^2du`

`=int(sqrt(2)sec^2(u))/(2^2(tan^2(u)+1)^2)du`

Take the constant out,

`=sqrt(2)/2^2int(sec^2(u))/(tan^2(u)+1)^2du`

Use the identity:`1+tan^2(x)=sec^2(x)`

`=sqrt(2)/4int(sec^2(u))/(sec^2(u))^2du`

`=sqrt(2)/4int1/(sec^2(u))du`

`=sqrt(2)/4intcos^2(u)du`

Use the trigonometric identity:`cos^2(x)=(1+cos(2x))/2`

`=sqrt(2)/4int1/2(1+cos(2u))du`

Take the constant out,

`=sqrt(2)/8int(1+cos(2u))du`

Apply the sum rule,

`=sqrt(2)/8{int1du+intcos(2u)du}`

Apply the common integral:`intcos(x)dx=sin(x)`

`=sqrt(2)/8{u+1/2sin(2u)}`

Substitute back `u=arctan(x/sqrt(2))`

`=sqrt(2)/8{arctan(x/sqrt(2))+1/2sin(2arctan(x/sqrt(2)))}`

`=sqrt(2)/8{arctan(x/sqrt(2))+1/2(2sin(arctan(x/sqrt(2)))cos(arctan(x/sqrt(2))))}`

`=sqrt(2)/8{arctan(x/sqrt(2))+sin(arctan(x/sqrt(2)))cos(arctan(x/sqrt(2)))}`

Use the identities:`sin(arctan(x))=x/sqrt(1+x^2),cos(arctan(x))=1/sqrt(1+x^2)`

`=sqrt(2)/8{arctan(x/sqrt(2))+x/sqrt(x^2+2)sqrt(2)/(sqrt(x^2+2))}`

`=sqrt(2)/8{arctan(x/sqrt(2))+(sqrt(2)x)/(x^2+2)}`

`=1/8{sqrt(2)arctan(x/sqrt(2))+2x/(x^2+2)}`

Add a constant C to the solution,

`=1/8(sqrt(2)arctan(x/sqrt(2))+(2x)/(x^2+2))+C`

 

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