Okay,
Can I write like this:
integral x^-4=x^-3/-3 ?
And one more question, from where is 8/9? In this exercise
http://www.part.lt/img/c1e69784c5724...e583256f86.jpg
Printable View
Okay,
Can I write like this:
integral x^-4=x^-3/-3 ?
And one more question, from where is 8/9? In this exercise
http://www.part.lt/img/c1e69784c5724...e583256f86.jpg
Yes, $\displaystyle \displaystyle \int{\frac{1}{x^4}\;{dx} = -\frac{1}{3x^3}+k$.Quote:
Originally Posted by Kristina
After differentiating, solve for $\displaystyle dx$ and multiply both sides by $\displaystyle 8x^2$:Quote:
And one more question, from where is 8/9? In this exercise
$\displaystyle \displaystyle u = 3x^3-1 \Rightarrow \frac{du}{dx} = 9x^2 \Rightarrow dx = \frac{1}{9x^2}\;{du} \Leftrightarrow 8x^2dx = \frac{8}{9}\;{du}. $
So,
1)-1/3x^3 fluxcion will be -3x^-3=9x^-4 I'm not right? (Wondering)
2)HOwever I have integration simalar like this example and I tried to use it. But I find problem
Integration (x^2-2) ln(x)dx
u=x^2-2
du=2x
du/dx=2x
dx=du/2x
ln(x)/2x*du
What i do wrong?
What are you doing here? trying to integrate? $\displaystyle \displaystyle \dfrac{-1}{3x^3} \neq -3x^{_3}$Quote:
Originally Posted by Kristina
It is written as $\displaystyle \displaystyle \dfrac{-1}{3x^3} = \dfrac{-x^{-3}}{3}$
Is your question $\displaystyle \displaystyle \int (x^2-2)\;ln(x)\;dx$???Quote:
Originally Posted by Kristina
Then start with $\displaystyle \displaystyle \int (x^2-2)\;ln(x)\;dx = \int x^2\;ln(x)\;dx - \int 2\;ln(x)\;dx$
Use integration by parts for the first term!
Okay,
Now I do not understand nothing... (Headbang)
integration u*du dx=(x^3/3-2x*1/x (CAN SOMEONE EXPLAIN HOW TO WRITE FORMULAS?)
If you posted more clearly and did not confuse your threads with new questions (which I have had to move several times now) you might have a better chance of understanding the help people are giving you.Quote:
Originally Posted by Kristina
Click on the link in my signature for how to format equations.
Thread closed.
