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

Question

745 views around the world

You can reuse this answer
Creative Commons License

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.!