Question #5ce2d

Question

1291 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-10T11:49:59

See below.

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

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

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

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