CatchThis2
Junior Member
- Joined
- Feb 6, 2010
- Messages
- 96
(5/x^6)- 4 to the third root of x^2
So far I made u=x^2
du/dx=1
du=dx
Where do I go from here?
So far I made u=x^2
du/dx=1
du=dx
Where do I go from here?
CatchThis2 said:(5/x^6)- 4 to the third root of x^2
So far I made u=x^2
du/dx=1<<< Incorrect du/dx = 2x
du=dx
Where do I go from here?
CatchThis2 said:No , it is this.
Evaluate the integral: 5 divided by x to the 6, minus 4 to the cubed root of x squared.
CatchThis2 said:This is what it should look like
Subhotosh Khan said:CatchThis2 said:This is what it should look like
CatchThis2,
use grouping symbols, and then you may interpret it as one of the acceptable forms
that Subhotosh Khan's rewrite showed:
\(\displaystyle \int\bigg(\frac{5}{x^6} \ - \ 4\sqrt[3]{x^2}\bigg)dx =\)
\(\displaystyle 5\cdot \int x^{-6} dx \ - \ 4\cdot \int x^{\frac{2}{3}} dx\)
CatchThis2 said:So now I have 5 x^-7/-7 minus 4 x^2/3
\(\displaystyle . . . You \ \ didn't \ \ find \ \ the \ \ correct \ \ antiderivatives,\)
\(\displaystyle \ \ and \ \ you \ \ left \ \ off \ \ the \ \ "+ C."\)
Substitute in so I should have -5/7 x^-7 minus 4^2/3 . . . \(\displaystyle This \ is \ not \ correct.\)