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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 = x^2 + 1, 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|x^2 + 1| + C

However, we can simplify this further by using the inverse trigonometric function arctan. Recall that tan(arctan(x)) = x, so we have:

tan(arctan(x)) = x

1/(1 + tan^2(arctan(x))) = 1/(1 + x^2)

1/(1 + x^2) = cos^2(arctan(x)) = 1/(1 + x^2)

Therefore, we can write:

∫ 1/(1+x^2) dx = ∫ cos^2(arctan(x)) dx

= ∫ (1 + tan^2(arctan(x))) / (1 + x^2) dx

= ∫ (1 + x^2) / (1 + x^2) dx + ∫ tan^2(arctan(x)) / (1 + x^2) dx

= x + ∫ (1 - 1/(1 + x^2)) dx

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

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

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