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

Answer 1

#1/4arctan^2(x/2)+C#

#intarctan(x/2)/(4+x^2)dx#
We can use substitution. Let #u=arctan(x/2)#. Taking the derivative, we see that #du=(1/2)/(1+(x/2)^2)dx=(1/2)/((4+x^2)/4)dx=2/(4+x^2)dx#.

Thus:

#=1/2int(2arctan(x/2))/(4+x^2)dx=1/2intudu=1/2(u^2/2)=u^2/4=1/4arctan^2(x/2)+C#
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Answer 2

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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Answer from HIX Tutor

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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