Natural Log intergrals

[Sevastra]

Okay, I only have a few questions, I hope they can be all answered, if not.. a few would be great! Okay..

1) Find the indefinite integral: "intergal" of (x^3-3x+2)/(x-3) dx
So far on this one, I did long division, but I don't quite get the long division and how it works (that would be a great thing to explain to me), and I couldn't get past the long division, I know what to do after that.
2) Same thing as number one, couldn't get past division:
"intergal" of (x^4 +x-4)/(x^2+2) dx

3)These next 2 I just got plain stuck on.. Directions: find the indefinite integral by u-substitution, (let u be the denominator of the integrand)

"intergral" of (1)/(1+sqrt2x), so far I got u to = 1+ sqrt2x which is 1+2x^(1/2) and then du=2x^(-1/2) dx... thats where I got stuck, i just need help finishing it.. same with the next one..

4) "intergal of" sqrt x/(sqrt x)-3 dx

last one...
5) evaluate the definite integral
"intergal from" 4 to 0| (5)/(3x+1)
I didn't really get started on this one, I need help with the process, Sorry I asked so many questions, I hope someone can help me by Sunday, Thanks for your time

-Jenn

 

[galactus]

Forgot your algebra?.

        x^2+3x+6
     _________________________
x-3|x^3+0x^2-3x+2
       x^3 -3x^2
      __________________
            3x^2-3x+2
            3x^2-9x
            ________________
               6x+2
               6x-18
               _____________
                   20

\(\displaystyle \L\\x^{2}+3x+6+\frac{20}{x-3}\)

There. Is that a refresher?. Integrate away.

For #2, follow my lead from #1. Practice your long division.

You should be able to divide if you're in calc. Give it a go.


#3:

\(\displaystyle \L\\\int\frac{1}{1+\sqrt{2x}}dx\)

Let \(\displaystyle u=\sqrt{2x}, \;\ \frac{u^{2}}{2}=x, \;\ udu=dx\)

\(\displaystyle \L\\\int\frac{u}{1+u}du\)

Now, finish?.

#4:

\(\displaystyle \L\\\int\frac{\sqrt{x}}{\sqrt{x}-3}dx\)

\(\displaystyle \L\\=\int\frac{3}{\sqrt{x}-3}dx+\int{1}dx\)

For the first part:

Let \(\displaystyle u=\sqrt{x}, \;\ u^{2}=x, \;\ 2udu=dx\)

\(\displaystyle \L\\6\int\frac{u}{u-3}du\)

The second part is rather obvious.


#5:

\(\displaystyle \L\\\int\frac{5}{3x+1}dx\)

Let \(\displaystyle u=3x+1, \;\ du=3dx, \;\ \frac{du}{3}=dx\)

\(\displaystyle \L\\\frac{5}{3}\int\frac{1}{u}du\)

 

[Sevastra]

Thank you for answering all my questions so quickly, it helped me out alot. Yeah, I havn't seen long division in a long time, so I'm practicing, thanks alot