Question

Question #5ce2d

Answer 1

See below.

Explanation:

`2x^3+6x^2-x+4=2(x+alpha)((x+beta)^2+gamma^2)`

because any odd degree polynomial has at least one real root.

so

`1/(2(x+alpha)((x+beta)^2+gamma^2)) = A/(x+alpha)+(B x+ C)/((x+beta)^2+gamma^2)`

Now

`int(A/(x+alpha)+(B x+ C)/((x+beta)^2+gamma^2))dx=`

`=A log_e(x+alpha)+((C-beta B)/gamma)arctan((beta+x)/gamma)+1/2Blog_e((x+beta)^2+gamma^2)+C_0`