integration involving quadratic factor

N

nickname

New member
Joined
Mar 24, 2009
Messages
39
?(3x-1) / (x^2 - 3x + 4) dx

For this exercise, there is an ln and an arctan involved in the answer. I understand how to get the ln part, but could anyone please explain to me the arctan part of the answer? Thank you Is there is an easier method of solving the arctan I will grealty appreciate it.


?(3x-1) / (x^2 - 3x + 4) dx

(3x-1) / (x^2 - 3x + 4)= 3/2 (2x-1/3)(2) / (x^2 - 3x + 4)= 3/2 2x-3-2/3+3 / (x^2 - 3x + 4) =


?(3x-1) / (x^2 - 3x + 4) dx= 3/2 ?(2x-3)/(x^2 - 3x +4) dx + (3/2)(7/3) ?1/ (x-3/2)^2 - 9/4 + 4 =

3/2 ln (x^2 - 3x + 4) + (7/2)(2/ ? 7) arctan 2(x-3/4) / ? 7 + c

= 3/2 ln (x^2 - 3x + 4) + ? 7 arctan 2x-3 / ? 7 + c

Thanks again! :D
 
nickname said:
?(3x-1) / (x^2 - 3x + 4) dx

For this exercise, there is an ln and an arctan involved in the answer. I understand how to get the ln part, but could anyone please explain to me the arctan part of the answer? Thank you Is there is an easier method of solving the arctan I will grealty appreciate it.


?(3x-1) / (x^2 - 3x + 4) dx

(3x-1) / (x^2 - 3x + 4)= 3/2 (2x-1/3)(2) / (x^2 - 3x + 4)= 3/2 2x-3-2/3+3 / (x^2 - 3x + 4) =


?(3x-1) / (x^2 - 3x + 4) dx= 3/2 ?(2x-3)/(x^2 - 3x +4) dx + (3/2)(7/3) ?1/ (x-3/2)^2 - 9/4 + 4 =

3/2 ln (x^2 - 3x + 4) + (7/2)(2/ ? 7) arctan 2(x-3/4) / ? 7 + c

= 3/2 ln (x^2 - 3x + 4) + ? 7 arctan 2x-3 / ? 7 + c

Thanks again! :D
Click to expand...

That is a standard anti-derivative . I am not sure what you meant by explanation.

About "easier way" - there could be different way reach the same answer - and those may/may not be "deemed" easier.
 
Hello,

Thank you for replying. What I meant by explanation is that I don't completely understand how to solve the part of the problem that involves arctan. If anyone could explain me the steps toward solving the arctan part of the problem I will grealty appreciate it.

Thank you
 
\(\displaystyle tan^{-1}(x)=\int\frac{1}{x^{2}+1}dx\) is usually memorized and used as needed in lieu of deriving it each time it arises.

For your problem, say we complete the square in the denominator and make the sub \(\displaystyle u=x-\frac{3}{2}, \;\ du=dx\)

This gives:

\(\displaystyle 12\int\frac{u}{4u^{2}+7}du+22\int\frac{1}{4u^{2}+7}du\)

The right integral involves arctan and is \(\displaystyle \frac{\sqrt{7}}{14}tan^{-1}(\frac{2u}{\sqrt{7}})\)

Then, resub. I just used this to show arctan a little easier perhaps.

The left one involves ln.
 
\(\displaystyle After \ all \ is \ said \ and \ done, \we \ we \ get:\)

\(\displaystyle \int \frac{3x-1}{x^{2}-3x+4}dx \ = \ \frac{3}{2}ln|x^{2}-3x+4|+\sqrt7arctan\bigg(\frac{2x-3}{\sqrt7}\bigg)+C\)

\(\displaystyle A \ lot \ of \ grunt \ work, \ no \ matter \ how \ you \ slice \ it, \ unless \ one \ plugs \ the \ integral \ into \ his/her \ trusty\)

\(\displaystyle \ TI-89.\)
 
You must log in or register to reply here.
Top