# How do you find the integral of #int dx / (x²+4)²#?

I gave the wrong answer. Sorry!!! It has been too long since I did any Calculus. The correct answer is:

My original post deleted

I apologize once more. I need to do a self-study review of my calculus before answering any more questions of that kind. Solution verified in Maple.

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I found:

Here, we can utilize integration by parts to solve the problem as follows (I'll write it out for you if you can't tell; I don't want to write it too long):

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To find the integral of (\int \frac{dx}{(x^2 + 4)^2}), use the substitution method with a trigonometric identity:

- Let (x = 2\tan(\theta)), implying (dx = 2\sec^2(\theta) d\theta).
- Substitute (x = 2\tan(\theta)) into the integral: [ \int \frac{2\sec^2(\theta) d\theta}{(4\tan^2(\theta) + 4)^2} ]
- Simplify the denominator using the identity (\tan^2(\theta) + 1 = \sec^2(\theta)): [ \int \frac{2\sec^2(\theta) d\theta}{(4\sec^2(\theta))^2} = \int \frac{2\sec^2(\theta) d\theta}{16\sec^4(\theta)} ]
- Simplify further: [ \int \frac{1}{8} \sec^{-2}(\theta) d\theta = \frac{1}{8} \int \cos^2(\theta) d\theta ]
- Use the half-angle identity: (\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}): [ \frac{1}{8} \int \frac{1 + \cos(2\theta)}{2} d\theta ]
- Separate the integral and integrate: [ \frac{1}{16} \int (1 + \cos(2\theta)) d\theta = \frac{1}{16} \left( \theta + \frac{\sin(2\theta)}{2} \right) + C ]
- Substitute back for (\theta): Recall (x = 2\tan(\theta)), so (\theta = \tan^{-1}\left(\frac{x}{2}\right)). Also, (\sin(2\theta) = \frac{2\tan(\theta)}{1 + \tan^2(\theta)}) can be substituted as (\frac{2\tan(\theta)}{\sec^2(\theta)}) or (\frac{2\frac{x}{2}}{1 + \left(\frac{x}{2}\right)^2}), giving (\frac{x}{x^2/4 + 1}) after simplification.
- So, the integral becomes: [ \frac{1}{16} \left( \tan^{-1}\left(\frac{x}{2}\right) + \frac{x}{x^2 + 4} \right) + C ] Therefore, the integral of (\int \frac{dx}{(x^2 + 4)^2}) is (\frac{1}{16} \left( \tan^{-1}\left(\frac{x}{2}\right) + \frac{x}{x^2 + 4} \right) + 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.

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