# How do you find #int arctanx #?

We use the Integral of Inverse Functions theorem.

So

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The answer is

Execute this integral through part-by-part integration.

Consequently,

The essential component is

Lastly,

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Applying the Integration by Parts (IBP) Rule as follows:

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To find the integral of arctan(x), you can use integration by parts. Let u = arctan(x) and dv = dx. Then, du = 1/(1 + x^2) dx and v = x. Applying the integration by parts formula ∫u dv = uv - ∫v du, you get:

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

To integrate x/(1 + x^2), you can use the substitution method. Let u = 1 + x^2, then du = 2x dx. This simplifies the integral to ∫x/(1 + x^2) dx = ∫1/u du = ln|u| = ln(1 + x^2).

Therefore, the final result is:

∫arctan(x) dx = x arctan(x) - ln(1 + x^2) + C,

where C is the constant of integration.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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