Integration by parts MIDTERM really quick question

In summary, the conversation is about a question regarding integration by parts and specifically, the integral of arctan4t. The answer is t arctan4t-1/8 ln(1+16t^2)+C. The conversation also discusses a substitution of u=1+16t^2 and the confusion about using u again. Ultimately, the issue is resolved and the solution is found.
  • #1
banfill_89
47
0
integration by parts! MIDTERM...really quick question...please help!

Homework Statement



[tex]\int[/tex] arctan4t dt


Homework Equations





The Attempt at a Solution



ive been attempting this all day. the answer is t arctan4t-1/8 ln(1+16t^2)+C

i get this asnwer up until the 1/8...

a few steps befor emy answer i have t arctan4t - [tex]\int[/tex] 4t/ 1+16t^2

i don't kno whow to go from here to the answer...someone please help!
 
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  • #2


In your last integral, there's a fairly obvious substitution.

BTW, you need more parentheses: the integrand can be written more understandably as 4t/(1 + 16t^2)
 
  • #3


what would be the substituition? in the book after thsi they factor out a 1/8...im lost
 
  • #4


Substitute u=1+16t^2.
 
  • #5


? so am i integrating by parts again?
 
  • #6


banfill_89 said:
? so am i integrating by parts again?

No, it's a u substitution.
 
  • #7


i finnaly got it. this textbook sucks at explaining this. thanks guys. sorry dick i was confused at first cause u said let u= 16x^2+1 and i already used u. i get what you were saying now and it makes a lot of sense. thanks!
 

Related to Integration by parts MIDTERM really quick question

1. What is Integration by Parts?

Integration by Parts is a method used in calculus to find the integral of a product of two functions. It is based on the product rule of differentiation and is used to simplify integrals that are difficult to solve using other methods.

2. How do you use Integration by Parts?

To use Integration by Parts, you need to follow the formula: ∫udv = uv - ∫vdu, where u and v are two functions. You need to select u and dv in such a way that the integral on the right side of the formula is easier to solve than the original integral. Then, integrate the remaining integral on the right side using any other appropriate method.

3. When should I use Integration by Parts?

Integration by Parts is useful when the integral involves a product of two functions, one of which is difficult to integrate. It is also used when other methods, such as substitution, are not applicable.

4. What are the limitations of Integration by Parts?

Integration by Parts can only be used for definite integrals and cannot be used for indefinite integrals. Also, it may not always lead to a solution and may require multiple iterations of the formula to solve the integral.

5. Can Integration by Parts be used for all types of functions?

No, Integration by Parts is most effective for integrals involving logarithmic, trigonometric, and exponential functions. It may not work for other types of functions, such as rational functions or polynomials.

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