what is the integral of (dx)/[(x^2+2x+2)^2]?

Answer 1

HI again Mac;

∫ (dx)/[(x^2+2x+2)^2] =
∫1/ [(x^2+2x+2)^2] dx =
(1/2) [( x+1 / x^2 +2x +2) + tan^-1 (x+1)] +c

It is the Correct answer ;
Good Luck

Answer 2

This is int dx/((x+1)^2+1)
let x+1 =t

dx = dt
this is int 1/(t^2+1)dt = tan^-1 t or tan^-1 (x+1)

Answer 3

Homework: let x=tan(theta)-1, i.e. x+1=tan(theta).

You really should learn how to do these yourself. When the exam comes, you will be in a world of hurt otherwise.

Answer 4

dx/((x+1)^2 +1)^2

x+1= u, then dx=du

du / (u^2 + 1)^2

u = tgy.... du = (secy)^2dy

(secy)^2dy /((tgy)^2 + 1)^2 but (tgy)^2 + 1 = (secy)^2

(secy)^2dy/(secy)^4 = dy/(secy)^2 = (cosy)^2dy = (y + sin(2y)/2)/2 =( arctan(x+1) + sin2(arctan(x+1))/2)/2