Question

Can someone please help me out with this integral question? I tried everything that I could and I still can't figure it out..
Evaluate the integral.

(3x+2)/(x^2+0x–4) dx

*from 4 to 5

Thanks so much!

Answer 1

\[\frac{3x+2}{x^2-4}=\frac{\alpha}{x+2}+\frac{\beta}{x-2}\]
What are \(\alpha\) and \(\beta\)?
\[\int \frac{3x+2}{x^2-4}dx=\alpha\int\frac{1}{x+2}dx+\beta\int\frac{1}{x-2}dx\]
\[\alpha\int\frac{1}{x+2}dx+\beta\int\frac{1}{x-2}dx=\alpha \int \frac{1}{\gamma}d\gamma+\beta\int\frac{1}{\delta}d\delta\]
\[\alpha \int \frac{1}{\gamma}d\gamma+\beta\int\frac{1}{\delta}d\delta=\alpha \ln(\gamma)+\beta\ln(\delta)+C\]
Figuring out the Greek here will essentially explain the technique.

Answer 2

α=2 and β=1.. That's all I know and I don't know how to do the rest.. >_<

Answer 3

Ok. Let me explain.
You get \(\alpha\) and \(\beta\) via what's called "Heaviside Cover Up Method". You have an equation in \(x\), \(\alpha\), and \(\beta\). Observe:
\[\frac{3x+2}{x^2-4}=\frac{\alpha(x-2)+\beta(x+2)}{(x+2)(x-2)}=\frac{\alpha(x-2)+\beta(x+2)}{x^2-4}\]
Thus, we have:
\(3x+2=\alpha(x-2)+\beta(x+2)\)
Put in \(x=2\). This gives you: \(3\cdot 2+2=\alpha(2-2)+\beta(2+2)=4\beta\)
\(4\beta=3\cdot 2+2=8\). Thus, \(\beta\) is \(2\).
Put in \(x=-2\). This gives you \(3\cdot(-2)+2=\alpha(-2-2)+\beta(-2+2)=-4\alpha\).
\(-4\alpha=3\cdot (-2)+2=-4\). Thus, \(\alpha\) is \(1\).
So, we can thus say:
\[\int \frac{3x+2}{x^2-4}dx=\int\frac{1}{x+2}dx+2\int\frac{1}{x-2}dx\]
The next step is called "u-substitution". We know that the integral \(\int \frac{1}{u}du=\ln(u)\). In other words, it has a simple closed form. So, we let \(\gamma\) be \(x+2\) and \(\delta\) be \(x-2\). (Careful here. You can only do this if \(\frac{d\gamma}{dx}=1\) and \(\frac{d\delta}{dx}=1\). If you want clarification on precisely why, ask.)

This allows us to rewrite the original integral:
\[\int\frac{1}{x+2}dx+2\int\frac{1}{x-2}dx=\int\frac{1}{\gamma}d\gamma+2\int\frac{1}{\delta}d\delta\]
Since we can now integrate both of these, we have:
\[\int\frac{1}{\gamma}d\gamma+2\int\frac{1}{\delta}d\delta=\ln(\gamma)+2\ln(\delta)+C\]

Recall what our substitutions are and we can say:

\[\int\frac{3x+2}{x^2-4}dx=\ln(x+2)+2\ln(x-2)+C\]

Now here's your exercise. To evaluate a definite integral from \(a\) to \(b\), one must evaluate its indefinite integral at \(b\) and subtract from this its indefinite integral at \(a\). In other words,
\[\int_{a}^{b}f(x)=\left[\int f(x) dx\right]^{b}_{a}=[F(x)]^{b}_{a}=F(b)-F(a)\]

Use this to evaluate the indefinite integral: \[\int_{4}^{5}\frac{3x+2}{x^2-4}dx\]

Answer 4

Wow..You explained it so well! Thank you so much for your help!

Answer 5

You're welcome! I always have to be this exact whenever I learn things. I was hoping it would help you learn, too. :) You may find it interesting to look into precisely why we use the particular techniques for integration that we do use. It all has to do with finding a set of steps so that we can arrive at a "closed form." See ( http://en.wikipedia.org/wiki/Antiderivative) and, in general, ( http://en.wikipedia.org/wiki/Integral). There is an entire branch of mathematics dedicated to analyzing the structure of the group of closed form integrals. :)

Answer 6

I realized after doing what you did that I was doing it backwards the whole time.. Instead of ln(x+2)+2ln(x-2), I was using 2ln(x+2)+ln(x-2)... >_<

Answer 7

I see.. Thanks so much for your help!