MHB What is the result of integrating \frac{x}{x^2 + 1} and applying the limit?

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The integral of \(\frac{x}{x^2 + 1}\) from 1 to infinity can be solved using the substitution \(u = x^2 + 1\), leading to \(du = 2x \, dx\) and \(\frac{du}{2} = x \, dx\). This transforms the integral into \(\frac{1}{2} \int \frac{1}{u} \, du\), which evaluates to \(\frac{1}{2} \ln|u|\). Applying the limits from 1 to infinity results in \(\frac{1}{2} (\ln(\infty) - \ln(2))\), which diverges to infinity. Thus, the integral diverges, confirming that the limit does not yield a finite result. The conclusion is that the integral does not converge.
shamieh
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Suppose:

$$\int^{\infty}_1 \frac{x}{x^2 + 1} \, dx$$

I can't do this can I?

Let $$u = x^2 + 1$$
so $$du = 2x \, dx$$
$$\therefore \frac{du}{2} = x \, dx$$ ?
 
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That looks good to me! What do you get for the result of the integral? What happens when you then apply the limit?
 
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