Integration By Parts: Solving int.arctan(2x)dx for Calculus Homework

In summary, the conversation discusses using integration by parts to solve the problem of finding the integral of arctan(2x) dx. The solution involves using a u-substitution and checking the work by differentiating the antiderivative. The final answer is found using a table for arctan(u) and ln(sqrt(1+u^2)).
  • #1
mike01
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0
Integration By Parts?

Homework Statement


int.arctan(2x)dx


Homework Equations


Integration By Parts


The Attempt at a Solution



In the attached image is the original problem with the ansewer I came up with using integration by parts and then a v=sub. later in the problem I did not want to post additional steps because it turned out to be a longer problem than I thought Just curious if someone could confirm my ansewer and if it is incorrect I will post the work to help see where I messed up. thanks a ton.
 

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  • #2


I differentiated your antiderivative and I didn't get arctan(2x). If you want to check your work in the future, you could try that too. You can often get a clue where you messed up by looking at that as well.
 
  • #3


thanks I will see if I can figure it out.
 
  • #4


Yeah i almost got the same thing, except for the (1/4) looks like just a u-sub

Integral of arctan(2x) dx... u=2x du=2dx dx=(1/2)du

so now we have (1/2) integ arctan(u) du

leave the (1/2) out in front as a constant and I got u*arctan(u)-ln(sqrt(1+u^2))

plug everything back in and i got x*arctan(2x)-(1/2)ln(sqrt(1+4x^2)) ... but I just used a table for arctan(u)
 
Last edited:

Related to Integration By Parts: Solving int.arctan(2x)dx for Calculus Homework

1. What is Integration By Parts?

Integration by parts is a technique used in calculus to find the integral of a product of two functions. It is based on the product rule of differentiation and involves breaking down an integral into two parts and applying a specific formula to solve it.

2. When should I use Integration By Parts?

Integration by parts is typically used when the integral involves a product of functions that cannot be easily integrated by other methods such as substitution or partial fractions. It is also useful when the integrand involves a polynomial multiplied by a trigonometric or exponential function.

3. What is the formula for Integration By Parts?

The formula for integration by parts is ∫u dv = uv - ∫v du, where u and v are the two functions being integrated. This formula is derived from the product rule of differentiation.

4. How do I choose which function to use as u and which as dv?

The choice of u and dv depends on the integrand and can vary in different problems. A common method is to use the "LIATE" rule, where u is chosen in the following order: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential. The function of u should be one that becomes simpler after differentiation.

5. Are there any limitations to Integration By Parts?

Yes, integration by parts may not be applicable to all integrals. In some cases, it may lead to an infinite loop, making it impossible to solve. It is important to recognize when integration by parts is not a suitable method and try alternative techniques.

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