How do you determine if the improper integral converges or diverges #int [ (x arctan x) / (1+x^2)^2 ] dx# from negative 0 to infinity?
Since "negative #0# " is #0# , we deal with the problem as #int_0^oo(xarctanx)/(1+x^2)^2dx#
Since "negative
Given Improper Integral converges to
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To determine if the improper integral ( \int_{\infty}^{0} \frac{x \arctan x}{(1 + x^2)^2} , dx ) converges or diverges, we need to analyze its behavior as ( x ) approaches negative infinity and as ( x ) approaches 0 from the left.

As ( x ) approaches negative infinity, we evaluate the limit: [ \lim_{a \to \infty} \int_{a}^{0} \frac{x \arctan x}{(1 + x^2)^2} , dx ]

As ( x ) approaches 0 from the left, we evaluate the limit: [ \lim_{b \to 0^} \int_{b}^{1} \frac{x \arctan x}{(1 + x^2)^2} , dx ]
If both limits exist and are finite, the integral converges. If at least one of the limits is infinite or does not exist, the integral diverges.
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To determine if the improper integral ( \int_{\infty}^{0} \frac{x \arctan(x)}{(1+x^2)^2} , dx ) converges or diverges, you need to analyze the behavior of the integrand as ( x ) approaches negative infinity and as ( x ) approaches 0 from the left.
First, consider the behavior of ( \frac{x \arctan(x)}{(1+x^2)^2} ) as ( x ) approaches negative infinity. As ( x ) goes to negative infinity, ( \arctan(x) ) approaches ( \frac{\pi}{2} ) and ( (1+x^2)^2 ) approaches infinity. Thus, ( \frac{x \arctan(x)}{(1+x^2)^2} ) approaches 0.
Next, consider the behavior of ( \frac{x \arctan(x)}{(1+x^2)^2} ) as ( x ) approaches 0 from the left. As ( x ) approaches 0 from the left, ( \arctan(x) ) approaches ( 0 ) and ( (1+x^2)^2 ) approaches ( 1^2 = 1 ). Thus, ( \frac{x \arctan(x)}{(1+x^2)^2} ) approaches 0.
Since the integrand approaches 0 as ( x ) approaches negative infinity and as ( x ) approaches 0 from the left, the integral may converge.
To determine convergence, integrate the function from negative infinity to 0. However, the given integral extends from negative infinity to positive infinity. So, evaluate the integral from 0 to infinity and then double the result.
Thus, the integral from 0 to infinity converges if the integral from negative infinity to 0 converges. Since the integral from 0 to infinity converges, the integral from negative infinity to 0 also converges.
Therefore, the improper integral ( \int_{\infty}^{0} \frac{x \arctan(x)}{(1+x^2)^2} , dx ) converges.
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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