(x ^ 2 + 1) / ((x + 2) ^ 2 (x ^ 2 + 2x + 3) ^ 2) indefinite integral

(x ^ 2 + 1) / ((x + 2) ^ 2 (x ^ 2 + 2x + 3) ^ 2) indefinite integral

(x + 2) / (x + 2) / (x ^ 2 + 2 + 2x + 3) DX = (x + 1 + 1) / (x ^ 2 + 2x + 3) DX (x + 2) / [(x + 2) / [(x + 2) / (x + 2) / (x ^ 2 + 2) / (x ^ 2 + 2 + 2 + 2x + 3) DX = (x + 1 + 1) / (x + 2 + 2 + 2x + 2 + 2x + 3 + 3) DX = 1 / 2 * ln (x (x (x ^ 2 + 2 + 2x2 + 2x + 2 + 2x + 2 + 2x + 2 + 2x + 2 + 2 + 2x + 2 + 2 + 2x + 2 + 2 + 2x + 2 + 3) + + 3 + 3 + 3) + 1 / 2 / 2 / 2 / 2 / 2 / 2 * as the as the as one / 2 / 2 / 2 * arctan (as the as the as the as the as the as the 3) +

Seeking indefinite integral 1 / x ^ 2 + 2x + 3

Let u = x + 1
Then Du = DX
The original formula = ∫ 1 / (2 + U & # 178; 2) Du
=1/√2·arctan(u/√2)+C
=1/√2·arctan[(x+1)/√2]+C

Indefinite integral ∫ {(2x-x ^ 2)} ^ (- 1 / 2) should be explained in detail
Indefinite integral ∫ {(2x-x ^ 2)} ^ (- 1 / 2) should be explained in detail

First, 2x-x ^ 2 = 1 - (x-1) & sup2;
Let X-1 = cost
It's a trigonometric integral
See resources for details