What is the Integral of # tan^-1x/(1+x^2)#?
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The integral of (\frac{\tan^{-1} x}{1+x^2}) with respect to (x) can be found using integration by parts. Let (u = \tan^{-1} x) and (dv = \frac{dx}{1+x^2}). Then, (du = \frac{1}{1+x^2} dx) and (v = \int \frac{dx}{1+x^2} = \tan^{-1} x).
Applying the integration by parts formula:
[ \int u , dv = uv - \int v , du ]
We get:
[ \int \frac{\tan^{-1} x}{1+x^2} , dx = \tan^{-1} x \cdot \tan^{-1} x - \int \tan^{-1} x \cdot \frac{1}{1+x^2} , dx ]
The remaining integral is a standard integral, which can be solved using substitution. Let (w = \tan^{-1} x), then (dw = \frac{1}{1+x^2} dx). Substituting these into the integral:
[ \int \tan^{-1} x \cdot \frac{1}{1+x^2} , dx = \int w , dw ]
Integrating (w , dw) with respect to (w):
[ \int w , dw = \frac{1}{2} w^2 + C = \frac{1}{2} (\tan^{-1} x)^2 + C ]
Therefore, the integral of (\frac{\tan^{-1} x}{1+x^2}) with respect to (x) is:
[ \int \frac{\tan^{-1} x}{1+x^2} , dx = \tan^{-1} x \cdot \tan^{-1} x - \frac{1}{2} (\tan^{-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.
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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