Indefinate Integrals

  • #1

Homework Statement


Hi, I'm having trouble finding the indefinite integral of an equation.


Homework Equations


The function is (2x-13)/[(x^2-2x+4)^0.5]


The Attempt at a Solution


I thought it was a good idea to let u=x^2-2x+4, hence du/dx= 2x-2.
From that I got [-11/(u^0.5)].du but evaluating this gives me -22(u^0.5), which I don't beleive is right.
Did I make a silly mistake or is my approach completely wrong? I am also having similar trouble finding the integral of [1+(1/t)]^5/(t^2).

Any help or advice would be very much appreciated, thank you.
 

Answers and Replies

  • #2
2,063
2
Your approach is correct, but you need rewrite the integral slightly.

[tex]\int{\frac{2x-13}{\sqrt{x^2-2x+4}}}[/tex]

[tex]\int{\frac{2x-2}{\sqrt{x^2-2x+4}}} + \int{\frac{-11dx}{\sqrt{x^2-2x+4}}}[/tex]

Now substitute [itex]u=x^2-2x+4[/itex], in the first integral.



As for the other integral, you can rewrite 1 + (1/t) as (t+1)/t.
 
Last edited:
  • #3
44
0
for your second q, i would manipulate it to look like (t+1)^5/t^7 and use binominal expansion for the top.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
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[tex]\int \frac{2x-13}{\sqrt{x^2- 2x+ 4}} dx= \int \frac{2x-2}{\sqrt{x^2- 2x+4}} dx- \int \frac{11}{\sqrt{x^2-2x+ 4}}dx[/tex]
Since the derivative of x2- 2x+ 4 is 2x- 2, the first can be done with the substitution u= x2-2x+ 4, but the second cannot.

Rewrite x2- 2x+ 4 as x2-2x+ 1+ 3= (x-1)2+3 (in other words, complete the square), let [itex]v= \sqrt{3}(x-1)[/itex] and then use a trig substitution
 
  • #5
2,063
2
and then use a trig substitution
Or as dexter would say, use a hyperbolic trig substitution. :biggrin:
 
  • #6
Thank you soo much for all the help guys.
One question about HallsofIvy's comment. I understand why and how you completed the square. But don't understand why v=sq.rt.3(x-1). Now dv/dx=sq.rt.3. If I substitute this into the denominator I get a function that I don't know what to do with.
I tried to use standard integrals and basically pulled the 11 in front of the integration sign and was left with 1/[(x-1)^2+(3^0.5)^2]^0.5. Which using standard integral formula I was left with arcsinh[(x-1)/(3^0.5)]+c.
Again I suspect I am wrong with this approach.
 
  • #7
2,063
2
(-11)arcsinh[(x-1)/(3^0.5)]+c is correct.
 
  • #8
Great. Thanks again everyone.
 

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