can some show me step by step please
thanks
j) Let me do j for you step by step.
$\displaystyle \int{\frac{{x}}{{x^2 + 2}}}dx$
Let $\displaystyle u = x^2 + 2$
Then $\displaystyle du = 2xdx$
Notice that our $\displaystyle du$ is the same as the numerator of our given except of the numerical coefficient $\displaystyle 2$. Do you see it?
Now, we can put the $\displaystyle 2$ by doing this.
$\displaystyle \frac{1}{2}\int{\frac{2xdx}{x^2+2}}$ Is this equal to the original given? Of course yes. So we can continue.
Then
$\displaystyle \frac{1}{2}\int{\frac{2xdx}{x^2+2}} = \frac{1}{2}\int{\frac{du}{u}}$.
There is a formula in integral calculus that states that the $\displaystyle \int{\frac{du}{u}} = \ln\left|u\right| + C$.
Hence in our problem..
$\displaystyle \frac{1}{2}\int{\frac{2xdx}{x^2+2}} = \frac{1}{2}\int{\frac{du}{u}} = \frac{1}{2}\ln\left|u\right| + C$.
Plugging in $\displaystyle u$.
$\displaystyle \int{\frac{{x}}{{x^2 + 2}}}dx = \frac{1}{2}\ln\left|x^2 + 2\right| + C $
You can check by getting the derivative of our final answer. If it results to the given, then we're pretty much correct.
I hope this helps.
For h, let me ask first, is it a Definite Integral? With limits 2 and 3? Do you know how to evaluate Definite Integrals?
The problem is quite easy and you can try it by yourself.
Try to continue this, just feel free to ask if you have some questions.
Hint: Start with this:
$\displaystyle 2\int_2^3 {d\theta + } \frac{1}{2}\int_2^3 {\cos 2\theta (2d\theta )}$
Hey darth! If I were you, I would change my signature. You should write the work function so that it doesn't make you look like you don't know that the limit of a constant is that constant.
I'm just joking with you. Sometimes people can't pick up on humor through text.
The limit as x tends to infinity of work is, guess what?
MORE WORK! Get it?