Indefinite integral (x + 2) / (x ^ 2 + 2x + 2)

Indefinite integral (x + 2) / (x ^ 2 + 2x + 2)

∫(x+2) / (x^2+2x+2) dx
=∫(x+1+1) / 1+(x+1)^2 dx
=∫(x+1) / 1+(x+1)^2 d(x+1) + ∫1 / 1+(1+x)^2 d(x+1)
=(1/2)ln[1+(x+1)^2] + arctan(1+x) + C

Calculating indefinite integral ∫ (2x + 3) ^ 3DX

Calculating indefinite integral ∫ (2x + 3) ^ 3DX
∫(2x+3)^3dx
= 1/2 * ∫(2x+3)^3d(2x+3)
= 1/2 * 1/4 * (2x+3)^4 + C
= 1/8 * (2x+3)^4 + C

Indefinite integral ∫ (2x + 1) ^ 3DX
Why?
1.dx=1/2d(2x+1）
2. Let DX = 1 / 2D (2U + 1) = 1 / 2du after u = 2x + 1

∫（2x+1)^3dx
=∫（8x^3+12x^2+6x+1）dx
=2x^4 + 4x^3 + 3x^2 + x +C
DF (x) is equivalent to the derivation of F (x), DF (x) = f '(x) DX
So D (2x + 1) = 2DX
dx=1/2d(2x+1）
du = d(2x+1) = 2dx
So DX = 1 / 2D (2U + 1) = 1 / 2du