# How do you integrate #( (arctan(x/2) ) ÷ (4+x²) dx#?

Thus:

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To integrate ( \frac{{\arctan\left(\frac{x}{2}\right)}}{{4+x^2}} ) with respect to ( x ), you can use substitution. Let ( u = \frac{x}{2} ). Then, ( du = \frac{1}{2} dx ). This gives us:

[ \int \frac{{\arctan\left(\frac{x}{2}\right)}}{{4+x^2}} dx = 2 \int \frac{{\arctan(u)}}{{4+(2u)^2}} du ]

Now, apply another substitution, let ( v = 2u ). Therefore, ( du = \frac{1}{2} dv ) and ( u = \frac{v}{2} ). Substituting these into the integral gives:

[ 2 \int \frac{{\arctan(u)}}{{4+(2u)^2}} du = \int \frac{{\arctan\left(\frac{v}{2}\right)}}{{4+v^2}} dv ]

This is a standard integral that can be solved using the properties of the arctan function. The integral evaluates to:

[ \frac{1}{4} \left( \pi \arctan\left(\frac{x}{2}\right) + 2\ln(4+x^2) \right) + 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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