How do I find the integral #intarctan(4x)dx# ?
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To find the integral of ( \int \arctan(4x) , dx ), you can use integration by parts. Let ( u = \arctan(4x) ) and ( dv = dx ). Then, ( du = \frac{1}{1+(4x)^2} , dx ) and ( v = x ).
Now, apply the integration by parts formula: [ \int u , dv = uv - \int v , du ]
Substitute the values of ( u ), ( dv ), ( du ), and ( v ) into the formula: [ \int \arctan(4x) , dx = x \arctan(4x) - \int \frac{x}{1+(4x)^2} , dx ]
The integral on the right side can be solved by substituting ( u = 1 + (4x)^2 ) and ( du = 8x , dx ): [ = x \arctan(4x) - \frac{1}{8} \int \frac{1}{u} , du ]
[ = x \arctan(4x) - \frac{1}{8} \ln|u| + C ]
Finally, substitute back ( u = 1 + (4x)^2 ) and simplify: [ = x \arctan(4x) - \frac{1}{8} \ln(1 + (4x)^2) + C ]
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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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