The correct answer to the integral ∫(1/((x+1)√(x^(2)+2x+2))) is In (|√(x^2 + 2x +2 - 1)|) - In (√(x^2 + 2x + 2)+1)/2 + C.
Partial Fractions: We can first decompose the integrand using partial fractions. This involves finding constants A and B such that:
1/((x+1)√(x^2+2x+2)) = A/(x+1) + B/√(x^2+2x+2)
Multiplying both sides by the denominator, we get:
1 = A√(x^2+2x+2) + B(x+1)
Substituting x=-1, we get A=1.
Substituting √(x^2+2x+2)=0, we get B=-1.
Therefore, the integrand can be rewritten as:
1/((x+1)√(x^2+2x+2)) = 1/(x+1) - √(x^2+2x+2)
Integration: Now, the integral can be easily solved using the following substitutions:
x+1 = t (dx=dt)
√(x^2+2x+2) = p (2x+2=dp/dx)
∫(1/((x+1)√(x^2+2x+2))) dx = ∫(1/t - p) dt
= In|t| - ∫p dt
= In|x+1| - ∫p dt
Integration of p: Integrating p requires a trigonometric substitution. Let:
x+1 = tanθ
dx = sec^2(θ) dθ
√(x^2+2x+2) = secθ
Substituting, we get:
∫p dt = ∫secθ dθ = In|secθ+tanθ| + C
Combining results: Substituting back the original variables, we obtain the final solution:
∫(1/((x+1)√(x^2+2x+2))) dx = In|x+1| - In|secθ+tanθ| + C
= In|x+1| - In|√(x^2+2x+2)+1| + C
= In|√(x^2+2x+2)-1| - In|√(x^2+2x+2)+1|/2 + C