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The integral of 1/(1 + x^2) dx is arctan(x) + C.

To evaluate the integral of 1/(1 + x^2) dx, we can use the substitution method. Let u = 1 + x^2, then du/dx = 2x and dx = du/2x. Substituting these into the integral, we get:

∫ 1/(1 + x^2) dx = ∫ 1/u * du/2x

= 1/2 ∫ 1/u du

= 1/2 ln|u| + C

= 1/2 ln|1 + x^2| + C

However, we can simplify this further by using the trigonometric substitution method. Let x = tanθ, then dx/dθ = sec^2θ and 1 + x^2 = 1 + tan^2θ = sec^2θ. Substituting these into the integral, we get:

∫ 1/(1 + x^2) dx = ∫ 1/(sec^2θ) * sec^2θ dθ

= ∫ cos^2θ dθ

= ∫ (1 + cos2θ)/2 dθ

= 1/2 ∫ 1 dθ + 1/2 ∫ cos2θ dθ

= 1/2 θ + 1/4 sin2θ + C

= 1/2 arctan(x) + 1/4 sin(2arctan(x)) + C

Since sin(2arctan(x)) = 2x/(1 + x^2), we can simplify the integral to:

∫ 1/(1 + x^2) dx = arctan(x) + C

Therefore, the integral of 1/(1 + x^2) dx is arctan(x) + C.

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