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Ok I have another integral problem. Equation below

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Well, the full solution is quite long. The general solution is that you have to use integration by parts: \[\int\limits_{}^{}udv = uv - \int\limits_{}^{}vdu\] Here, if you let u = arctan 4t, dv = dt you obtain \[du=\frac{ 4 }{16+t^{2}}dt\] and v = t. Then, you substitute back into the equation and solve. The integral of vdu should (if I'm doing my mental calculation correctly) be a u-sub for u as the denominator.
why wouldn't you make just arctan u?

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Because there is no integral formula for arctanu. We know the derivative of arctan u but the integral of arctan u must be done with integration by parts.
right but what i mean can we set u=arctan and dv=4t? or does the dt always have to be with one of the items you assign?
and wouldn't du above be \[\frac{ 4 }{ 1+16t ^{2} }\]
Well the thing is that u, v, du, and dv must all be functions. arctan by itself is not a function. What are you evaluating arctan at? You are right though. dv must be attached to dt. The purpose of integration by parts is to obtain all values u, v, du, and dv. if you had let u = dt, then how would you solve for u? If you integrate both sides, what are you integrating u with respect with? If you differentiate u, the derivative of dt has no meaning. And the answer is no to your question, we have u = arctan4t. If you treat the derivative as a fraction, you get \[\frac{ du }{ dt } = \frac{ 1 }{ 16+t ^{2} }\], then you must "multiply" both sides by dt. I must make the obligatory statement that du/dt is NOT NOT NOT NOT NOT a fraction. I will not be able to explain to you why this works without a treatment in differential forms which will confuse more than clarify but trust that it works. You can IMAGINE it as a fraction.
Perhaps your confusion about integration by parts arises from what it all means. The notation is highly obscure but you need to re-examine what it means. Ultimately, all the equations of u, v, and their derivatives are dependent upon the variable t. However, with the right choices, we can use u, v, du, and dv (remember! they are dependent upon t!!) to rearrange the integral into something much easier to solve. If you look at my original work, you will see I made the choice u = arctan 4t and dv = dt. u here is obviously a function of t. For dv = dt, if you integrate both sides you get v = t, another function of t. Ultimately, evaluating the right hand side of the equation will amount to \[t*\arctan(4t) - \int\limits_{}^{}\frac{ 4t }{ 16+t ^{2} }dt\] Wow. All these expressions are in terms of t. Now you just have to evaluate the integral with another proper method: u-substitution. Integration by parts is merely a way to rearrange integrals. It's the analogous integral version of the product rule of derivatives.
ok i still don't get how you're getting that = ( i know the derivative of arctan is 1/1+x^2 what is is the derivative of arctan(f(x))? isnt is 1/1+(f(x))^2 * f'(x)??
Your notation is slightly ambiguous: the derivative of arctan(f(x)) with respect to x is: (f'(x))/(1+f(x)^2). Here, f(t) = 4t. f'(t) = 4.
Oh I see what you're saying. My bad. I type slower than I think. It is 1 + 16t^2.
no thats how i meant i just put it at the end. usually i use the equation thingy but it takes me forever to post then but i meant for f'(x) to be in the numerator
If I'm understanding you correctly, you're absolutely right. I was in a rush to post a reply. The proper evaluation is \[t*\arctan(4t) - \int\limits_{}^{}\frac{ 4t }{ 1 + 16t ^{2} }dt\]
ok now what would i do from there with the next integral? i'm sorry I have so much trouble with these = (
would i use the 1/a srctan(x/a) + c?
or let u = 4tdt? = (
Now you do a u substitution. Why? because you noticed that the denominator is a power of 2 while the numerator is a power of 1. If we let u = 1 + 16t^2, we get du = 32 t dt. That means the integral is \[\frac{ 1 }{ 8 }\int\limits_{}^{}\frac{ du }{ u }\] which is obviously the ln|u| + c. Remember two things, first it's the natural log of the absolute value of u, and second, you are subtracting this integral. After that, you can re-substitute u = 1 + 16t^2 into the natural log and voila, you have the answer.
To your first question, no you cannot reduce it to the arctan formula. The arctan formula only works if your numerator is 1. To your second question, are you trying to do another integration by parts? That would be unwise since you can do a u-sub.
you make it sound so easy lol i know how to do almost all of the parts when dealing with integrating and derivatives but i just can't seem to put it all together = (((
ok are you up to help me with another one by chance?
No problem @bettyboop8904

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