∫ 1/(x²+1) dx = -1/(2i) ln(x-i) + 1/(2i) ln(x+i) + C.
Both methods lead to the same answer: ∫ 1/(x²+1) dx = tan^-1(x) + C.
Here's how to integrate 1/(x²+1):
Method 1: Trigonometric Substitution
Substitute: Let x = tan(θ). Then, dx = sec^2(θ) dθ.
Rewrite the integral: ∫ 1/(x²+1) dx = ∫ 1/(tan^2(θ)+1) sec^2(θ) dθ.
Simplify: Since tan^2(θ) + 1 = sec^2(θ), the integral becomes ∫ 1/sec^2(θ) sec^2(θ) dθ.
Integrate: ∫ 1 dθ = θ + C.
Substitute back: θ = tan^-1(x), so the integral becomes tan^-1(x) + C.
Therefore, ∫ 1/(x²+1) dx = tan^-1(x) + C.
Method 2: Partial Fractions
Factor the denominator: x²+1 cannot be factored using real numbers only.
Decompose into partial fractions: Assume 1/(x²+1) = A/(x+i) + B/(x-i), where A and B are constants to be determined.
Solve for A and B: Multiply both sides by the common denominator to get 1 = A(x-i) + B(x+i).
Set x = -i: This gives us 1 = -2Bi, so B = -1/2i.
Set x = i: This gives us 1 = 2Ai, so A = 1/2i.
Substitute back: 1/(x²+1) = 1/(2i)(x-i) + 1/(-2i)(x+i).
Integrate: ∫ 1/(x²+1) dx = 1/(2i) ∫ 1/(x-i) dx + 1/(-2i) ∫ 1/(x+i) dx.
Use the derivative of the inverse tangent function: ∫ 1/(x-i) dx = -i ln(x-i) + C1 and ∫ 1/(x+i) dx = i ln(x+i) + C2.
Combine the constants: C = C1 + C2.
Therefore, ∫ 1/(x²+1) dx = -1/(2i) ln(x-i) + 1/(2i) ln(x+i) + C.
Both methods lead to the same answer: ∫ 1/(x²+1) dx = tan^-1(x) + C.