lch7
- 17
- 0
So I was messing around with some basic u-sub calculus and came across this problem. Any help would be greatly appreciated!
[itex]\int[/itex] [itex]\frac{3x^{2}}{x^{2}-7}[/itex] dx
I'm using u-substitution, as stated at the beginning.
I see that the u could be the denominator and the numerator is the derivative, or du.
[itex]\int[/itex] [itex]\frac{du}{u}[/itex]
[itex]\int[/itex] [itex]\frac{1}{u}[/itex] du
ln|u|+C
ln|x[itex]^{2}[/itex]|+C
I'm pretty sure this is the correct answer, but I decided to integrate it with the boundaries x=1 and x=4.5.
[ln|x[itex]^{2}[/itex]-7|][itex]^{4.5}_{1}[/itex]
ln|13.25| - ln|-6|
ln(13.25) - ln(6)
2.5839-1.7917
.7922
So this is my final answer, but it just doesn't seem right. I graphed the function and saw it was under the x-axis. Is this the problem?
Homework Statement
[itex]\int[/itex] [itex]\frac{3x^{2}}{x^{2}-7}[/itex] dx
Homework Equations
I'm using u-substitution, as stated at the beginning.
The Attempt at a Solution
I see that the u could be the denominator and the numerator is the derivative, or du.
[itex]\int[/itex] [itex]\frac{du}{u}[/itex]
[itex]\int[/itex] [itex]\frac{1}{u}[/itex] du
ln|u|+C
ln|x[itex]^{2}[/itex]|+C
I'm pretty sure this is the correct answer, but I decided to integrate it with the boundaries x=1 and x=4.5.
[ln|x[itex]^{2}[/itex]-7|][itex]^{4.5}_{1}[/itex]
ln|13.25| - ln|-6|
ln(13.25) - ln(6)
2.5839-1.7917
.7922
So this is my final answer, but it just doesn't seem right. I graphed the function and saw it was under the x-axis. Is this the problem?