Integration by Parts substitution

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Homework Help Overview

The problem involves the integration of the arctangent function, specifically the integral \(\int \arctan(4t)dt\). Participants are exploring methods for solving this integral using integration by parts and substitution techniques.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply integration by parts but encounters difficulty with the resulting integral. They consider making a substitution for the denominator in the integral \(\int \frac{t}{1+16t^2}dt\). Another participant suggests letting \(u=1+16t^2\) and questions what \(du\) would equal in this context.

Discussion Status

Participants are actively discussing substitution methods and have provided some guidance on how to proceed with the integration. There is a recognition of the need for further clarification on the substitution process, but no consensus has been reached on the final steps.

Contextual Notes

There is an indication that the original poster is working within the constraints of a homework assignment, which may limit the types of solutions or methods they can employ. The discussion reflects uncertainty about the substitution process and its implications for the integral.

revolve
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Homework Statement



<br /> \int\arctan(4t)dt<br />

Homework Equations





The Attempt at a Solution



<br /> \int\arctan(4t)dt = t\arctan(4t) -4 \int \frac{t}{1+16t^2}dt<br />

I'm stuck at this point. I think I need to make a substitution for the denominator, but I'm not sure how to go about doing so.
 
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\int \frac{t}{1+16t^2}dt


f we let u=1+16t2 what is du equal to?
 
revolve said:

Homework Statement



<br /> \int\arctan(4t)dt<br />

Homework Equations





The Attempt at a Solution



<br /> \int\arctan(4t)dt = t\arctan(4t) -4 \int \frac{t}{1+16t^2}dt<br />

I'm stuck at this point. I think I need to make a substitution for the denominator, but I'm not sure how to go about doing so.

It's good so far.

You can substitute for V = {1+16t^2}, so dV = 32tdt.

Then, you'll have \frac{1}{32} \int \frac{1}{V}dV.

That equates to \frac {ln(|V|)}{32}.

Multiply by -4, to obtain \frac {-ln(|V|)}{8}.

Replace V and you're done.

Good luck,

Marc.
 
Got it. Thank you both! Very helpful.
 

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