i need some quick help with two integration problems

S

slinky2004

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Jan 18, 2005
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i've been trying to figure these out all day, but i cant figure these out:

*the capital S is my integration symbol, i don't know how to make a real one
1. S(3x^2 + x)/(3x - 2) dx
2. S(1)/(-3 + 4x - x^2)^1/2 dx

the problem i have is that i would normally solve these with substitution, but i cant find a U that will get me an acceptable dU. in problem 1, for example: if i pick U=3x^2 + x, my dU=6x + 1, if i pick U=3x - 2, my du=3 and i cant seem to do anything with either.
 
1. Long divide (or rewrite the numerator in terms of the denominator to decompose).

2. Complete the square on -x^2 + 4x - 3. If you don't recognise the resultant integrand, let sin(u) = x-2 and see for yourself.
 
i'm sorry, i dont understand the explanation of problem one can somebody give me some more detailed help with it? bump
 
The second problem can be a little tricky. Just a little.

Complete the square, as Unco suggested:

\(\displaystyle x^{2}-4x+3\)

\(\displaystyle x^{2}-4x=-3\)

\(\displaystyle x^{2}-4x+4=1\)

\(\displaystyle (x-2)^{2}-1\)

Multiply by -1:

\(\displaystyle 1-(x-2)^{2}\)

Letting u=x-2 and du=dx

\(\displaystyle \int\frac{1}{\sqrt{1-u^{2}}}du\)

Some may not think much of it, but I like trig sub:

Let \(\displaystyle u=sin{\theta}\) and \(\displaystyle du=cos{\theta}d{\theta}\)

\(\displaystyle \int\frac{cos{\theta}}{sqrt{1-sin^{2}({\theta})}}d{\theta}\)

Since \(\displaystyle 1-sin^{2}({\theta})=cos^{2}({\theta})\)

\(\displaystyle \int\frac{cos({\theta})}{sqrt{cos^{2}({\theta})}}d{\theta}\)

\(\displaystyle \int\frac{cos({\theta})}{cos({\theta})}d{\theta}\)

\(\displaystyle \int{1}d{\theta}={\theta}\)

Since \(\displaystyle {\theta}=sin^{-1}(u)\), we have

\(\displaystyle sin^{-1}(u)=sin^{-1}(x-2)\)

Now, you tackle the first one.
 
i've already got the second problem, but i cant figure out the first one. i cant find any way to rewrite the numerator so that it cancels with the denominator, long division and completing the square didn't help. can somebody help me? i've got class at 7:00 tonight. bump
 
Are you having trouble with the long dividing itself or the integrating after having long divided?
 
What do you mean by "cancelling the denominator"? By using long division, you should have gotten a polynomial plus a (much simpler) rational expression.

Please include the steps you have done when you reply. Thank you.

Eliz.
 
I believe that should be \(\displaystyle \frac{2}{3x-2}+x+1\). Watch those signs. Although they seem subtle, they can cause big error.

Now integrate. OK?.
 
We add the remainder over the divisor:

x + 1 + 2/(3x - 2).

Now you can integrate away.
 
i took the derivative of x, 1 and 2/3x-2 separately and got x^2/2 + x + 2/3ln(3x-2) is that right?
 
Did you get:

\(\displaystyle \frac{2ln(3x-2)}{3}+\frac{x^{2}}{2}+x\)?.

If you did, then :lol:
 
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