How to integrate this?

**Timbuk2**

[latex]
\displaystyle \int_2^6{\frac{1}{x^2 - 4x + 20}\,\,\,\,dx}
[/latex]

It seems simple enough but I can't for the life of me figure it out! I don't think a u substitution would work as [latex]du = (2x - 4)\,\,\,\,dx[/latex] which doesn't really help

Thanks!

LeixlipRed

Completing the square

Kevster

What if you put it to the power of -1? For example,

1/4x is the same as (4x)^-1

Kevin

Iwasfrozen

I'll give you a clue, the answer is 1/4(tan(1))

Now see if you can work that out yourself.

Iwasfrozen

Kevster wrote: »

What if you put it to the power of -1? For example,

1/4x is the same as (4x)^-1

Kevin

Here's another clue. Don't do this.

LeixlipRed

Kevster wrote: »

What if you put it to the power of -1? For example,

1/4x is the same as (4x)^-1

Kevin

While that is true it won't help you integrate it. You need to just rewrite the denominator by using the method of completing the square.

Kevster

That doesn't solve evenly though, or does it? the quadratic I mean

rebel girl 15

Kevster wrote: »

That doesn't solve evenly though, or does it? the quadratic I mean

By completing the square, wouldn't the bottom line read (x^2-4x+4)+16 which is (x-2)^2+4^2.

Substitute U for (x-2), then d(x-2)=du, dx=du.

so you then have 1 over u^2+4^2 and use standard integral which equals arctan(x/a)

I think thats the way to do it - I haven't done this integrating for a while!!

bnt

Iwasfrozen wrote: »

I'll give you a clue, the answer is 1/4(tan(1))

Now see if you can work that out yourself.

Can't say I agree with that answer... I get something else for the indefinite integral portion: [latex]\frac{1}{4}{atan}(\frac{x-2}{4})[/latex] (CAS solution).

From that I suspect it's going to involve the standard integral [latex]\int{\frac{1}{x^2 + 1}} dx = atan(x)[/latex], with a substitution? Not sure.

**Timbuk2**

Thanks for the help!

I managed to solve it with completing the square! The answer I got was
[latex]
\displaystyle \frac{1}{4}\tan^{-1}({\frac{x-2}{4}})
[/latex]
and subbing in the limits (2 to 6) I got [latex]\frac{\pi}{16}[/latex].

Thanks for the help!

bnt

Cool! Am I right in saying the indefinite portion went something like this?
[latex]\displaystyle \int{\frac{1}{x^2 - 4x + 20}\,dx}[/latex]

[latex]\displaystyle =\int{\frac{1}{(x^2 - 4x + 4) + 16}\,dx[/latex]

[latex]\displaystyle =\int{\frac{1}{(x-2)^2 +4^2}\,dx[/latex]

Then substitutions u for (x-2) and a for 4, and the standard integral:
[latex]\displaystyle \int\frac{dx}{u^2 + a^2} = \frac{1}{a}\tan^{-1}(\frac{u}{a}) +C[/latex].

I don't know how people remember all these in exams, I really don't ... I can do this stuff if I can just remember it. I use giac/xcas or my TI-89 to find the answers to these things, if possible.

rebel girl 15

I got it right! Happy out now, wasn't sure if I did it right or not! (I went as far as integrating, didn't put in the limits!)

bnt

The last part (definite integral) is pretty easy in this example, especially since the lower limit evaluates to zero: arctan (2-2) = arctan (0) = 0. With the upper limit, it's arctan (6-2)/4 = arctan 4/4 =arctan (1) = pi/4. The final answer is 1/4 that, so it's pi/16. (All Radians, of course, in case your calculator gives you weird results.)