Integrate x/(x^2-5x+25)

Question

When I substitute u=x^2-5x+25 I get du=2x-5 dx. That gets me close to x dx. I would normally just pull out 1/2, but I'm confused how to deal with the -5.  it looks like I should get a ln portion with the u-substitution and also an arctan portion. I'm missing something here.

Answer 1

∫xdx/(x²-5x+25)

The derivative of the denominator is 2x-5, alas the numerator is not that. 

∫xdx/(x²-5x+25)=(1/2)∫2xdx/(x²-5x+25) and now comes one of the neat secrets, add 0, as -5+5

∫xdx/(x²-5x+25)=(1/2)∫(2x-5+5)dx/(x²-5x+25)=(1/2)∫(2x-5)dx/(x²-5x+25)

+(1/2)∫5dx/(x²-5x+25)

The first of these is (1/2)log(x²-5x+25)

The second is (5/2)∫dx/((x-5/2)²+75/4)  complete the square for the denominator.  We almost have an arctan integral.

The second is (5/2)/(75/4)∫dx/[(4/75)(x-5/2)²+1)=(2/15)∫dx/[((2/√75)x-1/√3)²+1)

It helps to use 5√3 for √75

The second integral is (2/15)/(2/(5√3))∫(2/(5√3))dx/[((2/5√3)x-1/√3)²+1)=(1/√3)arctan[(2x-5)/(5√3)]

There has got to be an easier way to teach adding zero.

Answer 2

Judith,

My integral calculus is a little rusty, buuuuut:

1. 2x - 5 in the numerator would have been convenient. :)
2. With a constant over x² - 5x + 25, yes, you should be able to wrangle that into an arctan with messy constants.

Soooo, what to do.  Can you slightly alter the numerator so that you end up with both of those? That is, can you buy yourself a constant in the numerator to work with, bringing you closer to something that looks like 2x - 5, which you can take care of with #1?  And then, balance your new constant with another constant...which buys you something that looks like #2?    

 

Does that give you a bit of a nudge?

-- Michael

Answer 3

You can multiply the numerator and denominator by 2, then subtract 5 and add 5 to the numerator, then split into 2 integrals.

2x - 5 +5           dx                     2x-5        dx           +     5 dx

------------                                 ---------                         ------------

2(x^2-5x+25)                       2( x^2-5x+25)         2( x^2-5x+25) 

 

And you're right, first one involves Ln and second one involves inverse tangent.

 

Hope this helps.