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

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- Nov 14th 2010, 07:57 AMKristinaAnother integral.
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 - Nov 14th 2010, 10:54 AMTheCoffeeMachine
Yes, $\displaystyle \displaystyle \int{\frac{1}{x^4}\;{dx} = -\frac{1}{3x^3}+k$.

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}. $ - Nov 15th 2010, 03:33 AMKristina
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? - Nov 15th 2010, 05:06 AMharish21
What are you doing here? trying to integrate? $\displaystyle \displaystyle \dfrac{-1}{3x^3} \neq -3x^{_3}$

It is written as $\displaystyle \displaystyle \dfrac{-1}{3x^3} = \dfrac{-x^{-3}}{3}$

Is your question $\displaystyle \displaystyle \int (x^2-2)\;ln(x)\;dx$???

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! - Nov 16th 2010, 10:34 AMKristina
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?) - Nov 16th 2010, 12:01 PMmr fantastic
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.

Click on the link in my signature for how to format equations.

Thread closed.